# Software Engineering Interview Questions in San Jose, CA

Software engineering interview questions shared by candidates

## Top Interview Questions

### Software Engineer at Apple was asked...

You have a 100 coins laying flat on a table, each with a head side and a tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or in any other way find out which side is up. Split the coins into two piles such that there are the same number of heads in each pile. 36 AnswersAnswer #1: Place 50 coins into two piles on its edges so that both have the same amount of heads in each pile, neither facing up or down. Answer #2: Trick question, place 50 coins in both piles and in theory they all have heads just not necessarily facing up or down. agree with 2nd ans Split into two piles, one with 90 coins and the other with 10. Flip over every coin in the pile with 10 coins. Show More Responses Just split into two piles, each with 50 coins. The question only asks 50 heads in each one, it doesn't ask for the number of heads up!!! Pick 10 coins from the pile, flip it and put it in the other pile. This will ensure that the number of heads up are equal in both the piles Pick 10 coins from the original 100 and put them in a separate pile. Then flip those 10 coins over. The two piles are now guaranteed to have the same number of heads. For a general solution of N heads and a total of M coins: 1.) Pick any N coins out of the original group and form a second pile. 2.) Flip the new pile of N coins over. Done. Example (N=2, M=6): Original group is HHTTTT (mixed randomly). Pick any two of these and flip them over. There are only three possible scenarios: 1: The two coins you picked are both tails. New groups are {HHTT} {TT} and when you flip the 2nd group you have {HHTT} and {HH}. 2.) The two coins you picked consist of one head and one tail. New groups are {HTTT} and {HT} and when you flip the 2nd group you have {HTTT} and {TH}. 3.) The two coins you picked are both heads. New groups are {TTTT} and {HH} and when you flip the 2nd group you have {TTTT} and {TT}. The question says "'You' can't feel, see or in any other way find out which side is up....' Can a team member? Cooperate with a fellow engineer, or other colleague, who can see the coins to solve the problem? Question has its answer in it... 10 coins are head up..... 90 coins are tail down..... so it means all 90 coins are head up.... Now, all you have to do is to split it into half. 50/50 Let's generalise the question to where there are n heads and any number of tails on the table. Select any n coins. This set will contain m heads, where m is between 0 and n inclusive, and n - m tails. The other n - m heads will be in the remaining coins. We now have two piles: the selection of n coins with n-m tails and the remainder with n-m heads. All we have to do is flip the selection so that the n-m tails become n-m heads, the same number as the heads in the remainder. This is a straightforward extension of the 'pick any 10 coins and flip' answer correctly given above by several people. All of you are over thinking it. Read the last bloody line, "Split the coins into two piles such that there are the same number of heads in each pile" They're not asking for the heads to be up or down, just an equal amount & every coin has a head side so dividing the pile equally achieves that. 100 coins total, 10 of them are heads up, 90 are tails up. Meaning all of them are heads up AND tails down. Split it 50/50 and you are done. It is not as easy as to just split it. And it says heads UP tails UP. Given 10 h, 90 t. Pick some random 10 coins call it P1. Rest is P2. In P1, (10-x) heads, (x) tails In P2, (x) heads, (90-x) tails Flip the coins in P1. In P1, (x) heads and (10-x) tails P1 and P2 have the same number of heads. reading these answers is such a confidence builder. Show More Responses I agree to trev, don't think anyone read the question. we already have 2 piles --> 90 coins with tails up and 10 coins with heads up, just flip over 10 of the coins from 90 coins that have tails up, we will have same number of coins with heads up in each pile. get all coins in your hands, shake them, drop them. for each coin there is a 50% probability to lay heads up, and 50% probability for tails down. now split i half question doesn't need to look faces of which side is up after splitting it in two piles. split all coins in two part of 50 50 and they all have heads ...and thats what questioner asking..! and move them to the 10-coin pile. Take 40coins from 90-coin pile, flip them over and move to the 10-coin pile. It's really depends on whether Apple is hiring Software Engineers who are collaborators, mathematicians or tricksters. It's clear that Apple does hire Engineers who listen to the question accurately. Make two groups at random for 10 and 90 coins. Example:- G1(10) G2(90) case 1:- 6H,4T 4H,86T case2:- 3H,7T 7H,83T Now flip all coins of smaller group G1(10). The result will always have same Heads in each pile. G1(10) G2(90) case 1:- 6T,4H 4H,86T case2:- 3T,7H 7H,83T We just get 5 coins head up put in each piles ==> we get the same number of head up in each pile. They just ask we "Split the coins into two piles such that there are the same number of heads in each pile" . They didn't say that we don't kow what is coin head up and they mixed together. "The question says "'You' can't feel, see or in any other way find out which side is up....' Can a team member? Cooperate with a fellow engineer, or other colleague, who can see the coins to solve the problem?" This is the best answer yet! Completely out of the box answer and yet so simple. Show More Responses Flip every other coin, 90 Tails will get split into 45 Heads and 45 Tails. Similarly 10 Heads will get converted to 5 Head and 5 Tails, so now we have 50 heads (45 + 5) and 50 tails (45 + 5). Then just split them into two equal groups. Find complete answer with description here: http://www.puzzlevilla.com/puzzles/puzzle/172 Answer Make a pile of 10 and flip them over. Then the number of heads is equal in both piles. question says both group should have equal heads, but doesnt specifiy, it should be up, hence, just grouping 50 each would solve the problem This is a screw you question, but yeah if you take out 10 coins you can have anywhere between 0-10 heads for every head you have you have one less head in the other pile and one less tail in your pile of 10 coins. So if you have 100 coins 10 heads and you take lets say 10 coins 0 heads, 10 tails. The 90 coins has 10 heads. 1 heads, 9 tails. The 90 coins has 9 heads (you stole one when selecting 10 coins). 2 heads, 8 tails. The 90 coins has 8 heads (same you stole 2 when selecting 10 coins ect). 3 heads, 7 tails. The 90 coins has 7 heads. 4 heads, 6 tails. the 90 coins has 6 heads. 5 heads, 5 tails, the 90 coins has 5 heads. 6 heads, 4 tails, the 90 coins has 4 heads. 7 heads, 3 tails, the 90 coins has 3 heads. 8 heads, 2 tails, the 90 coins has 2 heads. 9 heads, 1 tails, the 90 coins has 1 heads. 10 heads, 0 tails, the 90 coins has 0 heads. As you can see whenever you take out 10 because your not only stealing from the pile of 90's heads your also offsetting the pile of 10 coins tails by 1 equally you have an equal connection between the tails you have in the pile of 10 coins as you do heads in the pile of 90 coins that your tails in 10 coins pile always equals heads in 90 coin pile. So you just flip over each coin in the pile of 10 coins and your tails becomes heads. So if you selected 1 head and in the 10 coins pile you had 9 heads in the 90 coins pile and 9 tails in the 10 coins pile, you are guaranteed after flipping each over once to have 9 heads in the 10 coins pile as tails becomes heads and 9 heads in the 90 coin pile, and ect, ect. This stands true for any pile that you know the amount of one category and 2 options, If you know you have 25 of one things, despite how many things there are if each thing had only two options like heads or tails, you know selecting 25 of them the same amount you know of one thing that when taking out 25 or the equal number of what you know of one thing is in there that what you unsucessfully try to filter out is the inverse of what you selected successfully to take out. Pick 10 coins, flip them and form a separate pile. The no.of tails in both pile will be equal inspite of your choice being a tails up coins or a heads up coins. Coz when u pick a tails up coin u r reducing the no.of tails up in the first pile and since u flip it its gonna b a heads up coin the second pile, if u r picking up a heads up a coin u turn it into a tails up coin in the second pile so that it can cancel out one tails up coin in the existing first pile. If it means heads up then separate the coins into one pile of 90 one pile of 10 then flip the ten coins it works with all scenarios Of sides you ended up choosing also like to point out that we can't feel them so we probably can't use our hands to flip them but I assume they would allow us to use something as how else would we separate them The answer lies in the exact wording of the question "Split the coins into two piles such that there are the same number of heads in each pile. " It does not specify heads need to be face up, so you would simply split the piles in 50 each and you have the same number of coins with heads in each pile. Take ten coins (consider as one pile, Pile A and other 90 coins as another pile, Pile B). Now you have two piles. Turn all coins as in pile A, you will end up with same number of heads in both piles. Ex: Scenario 1: Consider in Pile A, there are 2 heads and 8 tails. Hence in Pile B there will 8 heads.Now when you turn all the coins in Pile A you will end with 8 heads in Pile A. Hence both Pile A and Pile B have same number of heads. Scenario 2: Consider in Pile A, there are 10 heads. Hence in Pile B there will be 0 heads.Now when you turn all the coins in Pile A, you will end with 0 heads in Pile A. Hence both Pile A and Pile B have same number of heads. Scenario 3: Consider in Pile A, there are 0 heads. Hence in Pile B there will be 10 heads.Now when you turn all the coins in Pile A, you will end with 10 heads in Pile A. Hence both Pile A and Pile B have same number of heads. Show More Responses Take 10 coins.Split into two piles of 5 each.Flip all coins in one pile.Both piles now have equal heads and tails.Take another 10 and go through the same procedure.Follow the same process for the entire original pile.You end up with two sets of 5 piles having equal no. of heads and tails.Combine all 5 piles on each side and it's done. Its very simple. step 1 take group of 10 coins from all now flip this pile and you will get your answer. how? lets see cases 100 total ( 10 H + 90T) so you get group of 10 from them so lets assume you will get 4 h+6T , and (6H + 84T) then flip this smaller one new group will be 4T+ 6H so now we 2 groups 1 new 1 old 4t+6h and 6h+85T both have same number of heads .... LITERAL ANGLE Split 50/50. Both piles have the same number of heads. Parameters do not require each pile to have the same number of heads facing upward. TEAMWORK ANGLE Ask the most efficient, skilled coin identification analyst at Apple to identify the coins so the skilled sorting robot can separate the piles equally. PATRONIZING ANGLE Take a picture of the table with your iPhone and sending to a laborer hired to come sort for you via a services app in the app store. NEXT LEVEL QUANTUM ANGLE If the coins are in no way observable, the question is impossible to answer because the coins are sitting next to Schrodinger's cat and thus are in a state of both heads and tails until observed. |

### Senior Software Engineer at Facebook was asked...

Write some pseudo code to raise a number to a power. 11 Answerspretty trivial... int raise(num, power){ if(power==0) return 1; if(power==1) return num; return(raise(num, power-1)*num); } double Power(int x, int y) { double ret = 1; double power = x; while (y > 0) { if (y & 1) { ret *= power; } power *= power; y >>= 1; } return ret; } Show More Responses In Ruby: def power(base, power) product = 1 power.times do product *= base end product end puts "2^10 = 1024 = #{power(2,10)}" puts "2^0 = 1 = #{power(2,0)}" puts "2^1 = 2 = #{power(2,1)}" If I were an interviewer, I would ask the Aug 29, 2010 poster why he used bitwise operators, and whether he would deploy that code in a production environment, or if he merely wanted to demonstrate, for purposes of the interview, that he understands bitwise operations. Because it uses dynamic programming and is lots more efficient than your algorithm. If the power is not integer, use ln and Taylor series If I'm the interviewer, none of above answers is acceptable. What if y < 0? what if y < 0 and x == 0? I'm seeing an endless recursion that will eventually overflow the stack, and the none-recursive one just simply returns 1. There is a way to do this in a logN way rather than N. function power(x, n) { if n == 1 return x; // Even numbers else if (n%2 == 0) return square( power (x, n/2)); // Odd numbers else return power(x, n-1); } This is from Programming pearls.. interesting way. small mistake function power(x, n) { if n == 1 return x; // Even numbers else if (n%2 == 0) return square( power (x, n/2)); // Odd numbers else return power(x, n-1) * x; } # Solution for x ^ n with negative values of n as well. def square(x): return x * x def power(x, n): if x in (0, 1): return x if n == 0: return 1 if n < 0: x = 1.0 / x n = abs(n) # Even number if n % 2 == 0: return square(power(x, n/2)) # Odd number else: return x * power(x, n - 1) print ("0 ^ 0 = " + str(power(0, 0))) print ("0 ^ 1 = " + str(power(0, 1))) print ("10 ^ 0 = " + str(power(10, 0))) print ("2 ^ 2 = " + str(power(2, 2))) print ("2 ^ 3 = " + str(power(2, 3))) print ("3 ^ 3 = " + str(power(3, 3))) print ("2 ^ 8 = " + str(power(2, 8))) print ("2 ^ -1 = " + str(power(2, -1))) print ("2 ^ -2 = " + str(power(2, -2))) print ("2 ^ -8 = " + str(power(2, -8))) |

### Senior Software Engineer at Google was asked...

Given an array of numbers, replace each number with the product of all the numbers in the array except the number itself *without* using division. 8 AnswersO(size of array) time & space: First, realize that saying the element should be the product of all other numbers is like saying it is the product of all the numbers to the left, times the product of all the numbers to the right. This is the main idea. Call the original array A, with n elements. Index it with C notation, i.e. from A[0] to A[n - 1]. Create a new array B, also with n elements (can be uninitialized). Then, do this: Accumulator = 1 For i = 0 to n - 2: Accumulator *= A[i] B[i + 1] = Accumulator Accumulator = 1 For i = n - 1 down to 1: Accumulator *= A[i] B[i - 1] *= Accumulator Replace A with B It traverses A twice and executes 2n multiplicates, hence O(n) time It creates an array B with the same size as A, hence O(n) temporary space # A Python solution (requires Python 2.5 or higher): def mult(arr, num): return reduce(lambda x,y: x*y if y!=num else x, arr) arr = [mult(arr,i) for i in arr] # O(n^2) time, O(n) space Create two more arrays. One array contains the products of the elements going upward. That is, B[0] = A[0], B[1] = A[0] * A[1], B[2] = B[1] * A[2], and so on. The other array contains the products of the elements going down. That is, C[n] = A[n], C[n-1] = A[n] * A[n-1], and so on. Now A[i] is simply B[i-1] * C[i+1]. Show More Responses def without(numbers): lognums = [math.log10(n) for n in numbers] sumlogs = sum(lognums) return [math.pow(10, sumlogs-l) for l in lognums] Here are my 2 cents to do this in memory without creating temporary arrays. The simple solution , if division was allowed, was multiple all the elements of the array i.e. tolal = A[0]*A[1]]*....*A[n-1] now take a loop of array and update element i with A[i] = toal/A[i] Since division is not allowed we have to simulate it. If we say X*Y = Z, it means if X is added Y times it is equal to Z e.g. 2*3 = 6, which also means 2+2+2 = 6. This can be used in reverse to find how mach times X is added to get Z. Here is my C solution, which take pointer to array head A[0] and size of array as input void ArrayMult(int *A, int size) { int total= 1; for(int i=0; i< size; ++i) total *= A[i]; for(int i=0; i< size; ++i) { int temp = total; int cnt = 0; while(temp) { temp -=A[i]; cnt++; } A[i] = cnt; } } Speed in O(n) and space is O(1) #include #define NUM 10 int main() { int i, j = 0; long int val = 1; long A[NUM] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; // Store results in this so results do not interfere with multiplications long prod[NUM]; while(j < NUM) { for(i = 0; i < NUM; i++) { if(j != i) { val *= A[i]; } } prod[j] = val; i = 0; val = 1; j++; } for(i = 0; i < NUM; i++) printf("prod[%d]=%d\n", i, prod[i]); return 0; } void fill_array ( int* array, size ) { int i; int t1,t2; t1 = array[0]; array[0] = prod(1, size, array ); for(i = 1; i < size; i++){ t2 = array[i]; array[i] = prod(i, array.size(), array)*t1; t1 *= t2; } int prod(start, end, array){ int i; int val(1); for(i = start; i < end; i++ ) val *= array[i]; return val; } Hello, Thank you for sharing your interview experience. As a small team of ex-Google employees, we have recently launched a new website, interviewjoy.com, where you can earn money by sharing your interview experiences/insights with other job candidates. (It is a marketplace for sharing job interview insights). Posting an interview consultancy service is totally free & anonymous and we are giving 50 USD sign-up bonus for the first 500 users. You are kindly invited to interviewjoy.com to check it out. Users already started making money on the website! Best Regards.. (For more information: onboarding@interviewjoy.com) |

Suppose you have a matrix of numbers. How can you easily compute the sum of any rectangle (i.e. a range [row_start, row_end, col_start, col_end]) of those numbers? How would you code this? 8 AnswersIt can be done in constant time by precalculating sums of some basic rectangles (extending all the way to the border of the matrix). That precalculation times time O(n) by simple dynamic programming. Please elaborate, which "basic rectangles"? Are you recursively dividing each rectangle into 4 smaller rectangles? Precalc time for doing that is not O(n)?!? Compute the sum of the rectangles, for all i,j, bounded by (i,j), (i,m), (n,j), (n,m), where (n,m) is the size of the matrix M. Call that sum s(i,j). You can calculate s(i,j) by dynamic programming: s(i,j) = M(i,j) + s(i+1,j) + s(i,j+1) - s(i+1,j+1). And the sum of any rectangle can be computed from s(i,j). Show More Responses Awesome!! The answer is already popular in computer vision fields!! It is called integral imaging. See this page http://en.wikipedia.org/wiki/Haar-like_features It wasn't 100% clear to me, then I found the Wiki page http://en.wikipedia.org/wiki/Summed_area_table Let a[][] be the 2d array, int i=0; for( j = row_start; j <= row_end; j++) for( k = col_start; k <= col_end; k++) i+=a[j][k]; Iterate over matrix as an array storing (new sums array) in each position the cumulative sum up to that point. For each row in the desired submatrix we can compute its sum as a difference between its end and start positions. Repeat for other rows. Add up all the row sums. |

### Software Engineer In Test at Google was asked...

Implement a binary tree and explain it's function 4 AnswersBinary Search tree is a storage data structure that allows log(n) insertion time, log(n) search, given a balanced binary search tree. The following implementation assumes an integer bst. There's a million implementations. Just look on wikipedia for search and insert algorithms. Hi Xin Li, A binary tree is not the same as binary search tree.. A binary tree is a tree in which every node has atmost two children nodes. It is a k-ary tree in which k=2. A complete binary tree is a tree in which all nodes have the same depth. The fact is ttttttt t t. T to t. To. A a aaAs Sdsassss. Show More Responses Hello, Thank you for sharing your interview experience. As a small team of ex-Google employees, we have recently launched a new website, interviewjoy.com, where you can earn money by sharing your interview experiences/insights with other job candidates. (It is a marketplace for sharing job interview insights). Posting an interview consultancy service is totally free & anonymous and we are giving 50 USD sign-up bonus for the first 500 users. You are kindly invited to interviewjoy.com to check it out. Users already started making money on the website! Best Regards.. (For more information: onboarding@interviewjoy.com) |

Why does one use MSE as a measure of quality. What is the scientific/mathematical reason for the same? 2 AnswersMean-Square error is an error metric for measuring image or video quality it is popular video and image quality metric because the analysis and mathematics is easier with this L2-Norm metric. Most video and image quality experts will agree that MSE is not a very good measure of perceptual video and image quality. The mathematical reasoning behind the MSE is as follows: For any real applications, noise in the readings or the labels is inevitable. We generally assume this noise follows Gaussian distribution and this holds perfectly well for most of the real applications. Considering 'e' follows gaussian distribution in y=f(x) + e and calculating the MLE, we get MSE which is also L2 distance. Note: Assuming some other noise distribution may lead to other MLE estimate which will not be MSE. |

### Java Software Engineer at SAP was asked...

1.Java Basics. 2.Plethora of Multithreading questions.GC 3.Simple data structure. (BFS) variation 4.Database Basics 1 AnswerI fumbled in 1 questions which was the nail in the coffin. |

### Software Engineer at Facebook was asked...

You have two lightbulbs and a 100-storey building. You want to find the floor at which the bulbs will break when dropped. Find the floor using the least number of drops. 37 AnswersStart moving up in increments of 10 floors and dropping the bulb until it breaks (ie: drop from floor 10, if it doesn't break, drop from floor 20, etc.). Once the bulb breaks, move down to the floor above the last floor it broke on and start moving up floors in increments of one until the second bulb breaks. This results in a worst case scenario of 19 drops. Surely a binary search method would be more efficient i.e. starting at 50 and either going up or down 25 floors based on if it breaks or not. If you do a binary search, what happens if it breaks at floors 50 and 25? Show More Responses Do you know what a binary search is? You drop it from floor 12 next. If it breaks, you know it breaks between floors 12 and 1, inclusive. If it doesn't, you know it breaks between floors 13 and 25, inclusive. The main principle of binary search is that with each step you reduce your search space in half. Now your search space consists of only 12 floors. Wow, I want to get asked such a question in an interview! >>you drop it from floor 12 next... if you broke it on 50 and 25... you are out of luck and out of bulbs... 19 drops is not the best worst case scenario... imagine trying floor 16, if it breaks, you try 1 - 15 and thats 16 tries. if it doesn't break, then try floor 31 and if it breaks, then try 17 - 30 (so 16 tries, including the try on floor 16). And on and on (45, 58, 70, 81, 91, 100). If you reach 91, you'll have tried 7 floors so far and if it doesn't break, then there's 9 more tries to get to 100 (thus 16 in the worst case) Even a drop from the ceiling of 1st floor, a simple light bulb will break. thats what i think It's a light bulb. It will break when dropped from the 2nd floor. Drop it there then go to the first floor, hold it over your head and drop it. first do a binary search (agressive first step - fast) with 1 bulb. when first breaks, you know X(last but one fall - success) and Y(last fall - failure). now do a linear search starting from X(conservative but accurate second step - slow). complexity = in between logN and N. Use Binary search starting with the middle 50 The complexity of binary search is logN . So it will be log(100) < 7. Based on my experience, I think it will be floor 1 itself . Drop from 1st floor. If it didn't break, drop the same bulb from 2nd. If it still didn't break, drop the same bulb from 3rd... repeat as necessary. Only one light bulb required :) Yes, but doing each floor, that will give you the most drops -- question relates to optimizing for "least" number of drops -- I didn't think about being able to re-use the bulbs...that obviously is helpful. Maybe a fibonaci sequence once you determine a "break" floor and a "non-break" floor. I'd probably fail completely at coding it...knowledge of optimization and prediction theory would certainly be useful. Let f(m,k) be number of tries for m lamps and k floors. Obviously f(1,k)=k. let f(2,k) be s. k<=(s-1)+(s-2)...(1) =s(s-1)/2. Therefore f(2,100)=15. Show More Responses Actually, 16 is not the optimal, nor is 15; you can do it in 14. Here is one solution (there are at least 5 other equivalents): * Drop first bulb at 14th, 27th, 39th, 50th, 60th, 69th, 78th, 85th, 91st, 96th, and (optionally) 100th until it breaks. * Go to the highest floor where the first bulb didn't break. * Repeatedly go up one floor and drop the second bulb. When it breaks, you have your answer. Why is 14 optimal? Since you are decrementing each time, you want (n) such that sum(1..n) barely reaches 100. The answer is n=14. Generally, if the number of floors is (f), the lowest number of drops is ceil((sqrt(8*f+1)-1)/2). This is the best worst-case scenario. An interesting related question is what gives the lowest expected number of drops. And no, I could not have gotten this in an interview. In theory, use one bulb to determine an interval, and use the second bulb for a linear search within the interval. The solution that decreases the interval by one for each trial is quite clever. In practice, however, take the nature of the problem into account: Start on the first floor and move up by one floor. That's the answer I would be looking for. Start with bulb 1 and drop it from floor 2. Doesnt break? then floor 4 Doesnt break? keep dropping the same bulb moving even number of floors all the way upto floor 100. If on some even floor the bulb breaks drop the second bulb from the odd floor below the even floor, to detect if its the even or the odd floor that breaks the bulb Best case detection: 2 tries (first bulb breaks on 2nd floor, second bulb breaks on 1st floor) Worst case: 51 tries (the fist bulb breaks at 100 and second bulb breaks or does not break at 99th floor.. ) Go to the 100th floor and drop the first bulb. It WILL break. Done, 1 drop. It doesnt say whats the lowest floor it will break at, just at what floor will it break with least drops. Thus floor 100. Alright guys...you have two light bulbs. ...the second one has to be used for linear search, let the worst case number of items to be searched be x, then your interval will also have to be x, which will result a worst case of 100/x drops for the first light bulb. So now you are just trying to minimize n=100/x+x, find n'=0 when x=19...the candidate's answer is correct. I meant...x=10. and n=19. 0 drops, 1 bulb......stop thinking like computer nerds. Use physics or an engineering mindset. They didn't prohibit the use of any tools. Grab a scale, figure out force req'd to fracture bulb. Calculate acceleration due to gravity adjusting for air resistance/barometric pressure at location (trivial for anyone who took a 1yr physics course). Figure out how many meters up you need to be based on the req'd acceleration. Done.... @Rich: I am sure they were hoping for you to give them a computing answer since they don't do physics, and rather do computer science. mer's answer is correct: 14. Let F(s, n) be the optimal worst-case number drops for s stories and n bulbs. Suppose we drop at floor f, constituting one drop. If it breaks, we must make up to F(f-1, n-1) drops. If it doesn't break, we must make up to F(s-f, n) drops. Thus, for a given floor f, we have up to 1 + max { F(f-1, n-1), F(s-f, n) } drops. Optimizing over f: F(s, n) = 1 + min{ max { F(f-1, n-1), F(s-f, n) } for f in 1..s} Using the appropriate base cases, we have F(100, 2) = 14, as mer has given. Another way to think about it is in terms of decision trees. We want to construct a binary decision tree with leaf nodes for floors 1..100, and at most one "left" turn per path from root to leaf. To minimize the height of the tree, we want to reduce the variance in the length of each root-to-leaf path. This suggest we try dropping the first bulb at floors of the form: a, a-1, a-2, .. a-b, where the sum a + (a-1) + .. + (a-b) is equal to or barely over 100, so that determining almost any floor (possibly excluding the top few) takes a drops. Using this approach, we get the sequence of drops that mer has suggested. Well done @mer I have seen this question posed many ways, and that is the best answer I have ever seen. Sure hope I get asked this one now Show More Responses 14 In my experience light bulbs break when dropped from any height above 3 feet nice explanation from http://www.programmerinterview.com/index.php/puzzles/2-eggs-100-floors-puzzle/ Depends on how accurate u want to be. If i want exact answer, drop one from fifty, if it breaks start from first floor woth the remaining bulb. If it does not break, then start from fifty first florr. u will iterate fifty times as worst case. If u want a approximate answer, u can do binary way with give or take twenty five floors. Step over based on accuracy needed. You are all ignoring valuable information in this question. We are talking lightbulb, not bowling ball, and building, not step ladder. The bulb will almost certainly break by being dropped from the second floor (assuming US numbering conventions). The chance of it surviving a 3rd floor drop are miniscule, but not zero. The chance of a 4th floor drop, even less. Therefore, drop it from the 3rd floor first. If it breaks, go to the second floor and try. If that breaks you know the answer is 2. If it by some miracle doesn't break from the 3rd floor drop, the answer is 4, but take the elevator up one floor and drop it just to see. Rinse and repeat to 5, but since it will have already broken, go out and grab a beer, and tell your friends how much fun you just had. n*(n+1)/2 = 100. n approx = 14. In worst case u can figure it out in 14 drops. 14th, 27th, 39th, 50th, 60th, 69th, 78th, 85th, 91st, 96th, and (optionally) 100th until it breaks. I believe the number sequence should be: 14, 27, 39, 50, 60, 69, ** 77, 84, 90, 95, 99 **. The 9 floor gap between floor 69 and 78 would result in 8+8 = 16 drops worst case. Easy. Answer is zero. You don't need a test to find out that a lightbulb is going to break, even when you drop it from the first floor, because it's made of glass. BigO(100/X + X-1), Where X is the number of floors. 100/X calculates the dropping times to break the first one and X-1 is the additional maximum overhead to break the second one starting from the previous dropping floor to the floor the previous bulb was broken. If you solve the derivative of the above equation equal to zero, the optimal solution becomes 9.9 ~= 10 . Worst case = 100/10 + 10 -1 = 19 If its a glass bulb it will break from a 2ft height...i wont even care climbing any floors to check. Show More Responses Once you break the first light bulb, you are FORCED to start at the highest safe floor + 1 (i.e. highest floor that you safely dropped bulb #1 from) and drop bulb #2 one floor at a time, moving upward. Therefore, the algorithm for the first light bulb is critical (cuz once it breaks you are counting by 1's upward). Binary search starts at the 50th floor ==> max # drops = 50 Choosing fixed increments rather than binary search, e.g. start at 10, increment by 10, yields better results ==> 25 The ideal solution is 14 drops ==> Start at 14, increment by 14 each step until it breaks (leaving for the reader why 14 is optimal). It doesn't matter what floor you are on to make a bulb break. Doesn't it matter how high off the floor the bulb is dropped. If I am on the 5th floor and the bulb is sitting on the floor of the 5th floor, how high off of that 5th floor do I need to drop it before it breaks. This is a misleading question. The question doesn't say that you will drop the bulb out the window. Drop both from eye level. Both will break, and I answered the question without even climbing one stair. Efficiency isn't always about math..Common sense Answer: 14 drops Mathematically: 14 is the first number n, where the sum of numbers from 1 to n is greater than 100 Trial and error: The worst case happens when the bulbs break at floor 13. If you start from the 14th floor and the bulb breaks, then you start at the bottom floor and work your way up. If it doesn’t break and you try it again from the 28th floor and it breaks, then you go back down to the 15th and work your way up 1 floor at a time. |

### Software QA Engineer at Apple was asked...

There are three boxes, one contains only apples, one contains only oranges, and one contains both apples and oranges. The boxes have been incorrectly labeled such that no label identifies the actual contents of the box it labels. Opening just one box, and without looking in the box, you take out one piece of fruit. By looking at the fruit, how can you immediately label all of the boxes correctly? 41 AnswersAll the three boxes are names incorrectly. SO the bax lebeled Apples+Oranges contains only Oranges or Only Apples. Pick one fruit from it. If it is Orange then lebel the box as Orange. So the box lebeled Oranges contains Apples and the remaining contains both. Label the boxes fruit. The key bit is "All the three boxes are names incorrectly" so the label on the box which fruit comes from will need to be changes to one of the other 2 labels. It can only be 1 of them (and it will be obvious when you have the fruit) then the remaining box (that hasnt featured yet)...Just swap that label with fruit box that was originally on the box which you took the fruit out of Thats hard for anybody to understand somebody elses explanation... eaiest way is to just do an example Show More Responses Swaz answer is almost correct however it does not work in all scenarios. lets assume: box 1 is labelled Oranges (O) box 2 is labelled Apples (A) box 3 is labelled Apples and Oranges (A+O) and that ALL THREE BOXES ARE LABELLED INCORRECTLY" Pick a fruit from box 1, 1) if you pick an Orange: - box 1's real label can only be O or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - box 1's new label should then be A+O by elimination - since ALL LABELS ARE INCORRECT - box 2's label is changed to O - box 3's label is changed to A - SOLVED 2) if you pick an Apple: - box 1's real label can only be A or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - this still leaves us with the choice between label A and label A+O - which would both be correct - FAILURE Solution: The trick is to actually pick a fruit from the A+O labeled box Pick a fruit from box 3: 1) if you pick an Orange: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be O by elimination - since ALL LABELS ARE INCORRECT - box 1's label is changed to A - box 2's label is changed to A+O - SOLVED 2) if you pick an Apple: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be A by elimination (not O) - since ALL LABELS ARE INCORRECT - box 1's label is changed to A+O - box 2's label is changed to O - SOLVED it only says you can't look, doesn't mean you can't feel around or smell the fruit you picked, easy deduction after you figure the first box out Sagmi is right, but did not give the full reasoning. "the bax lebeled Apples+Oranges contains only Oranges or Only Apples. Pick one fruit from it. If it is Orange then lebel the box as Orange." so far so good Now, the box labelled Apples cannot be the box containing only Oranges, you've just found that box, so it must contain Apples and Oranges. And in that case the other box, labelled Oranges, must contain only Apples. It's easier to draw it out. There are only 2 possible combinations when all labels are tagged incorrectly. All you need to do is pick one fruit from the one marked "Apples + Oranges". If it's Apple, then change "Apple + Orange" to "Apple" The "Apple" one change to "Orange" The "Orange one change to "Apple + Orange" If it's Orange, then change "Apple + Orange" to "Orange" The "Apple" one change to "Apple + Orange" The "Orange" one change to ""Apple" Since all 3 boxes are labled incorectly Start with the box Labled A&O. If Its apples than the box labled apples then the apple one is oranges and the oranges is O&A. Label each box "Apples and/or Oranges" and the all will be correct. This is very simple to resolve. I was asked the same question at FileMaker. Each box is incorrectly labeled. So you go to the box that is labeled "Oranges and Apples" and take one out. It doesn't matter what comes out because all that you know is that it is not AO. If you remove an Apple then move the Apple label to it. Since the Apples are already identified it is easy to resolve the rest. All you know for certain is that the other two boxes remaining are mislabeled. So the AO label goes on the box with the remaining label and that label goes on the Apple box as you have already assigned that. The end result is you only need to remove one piece of fruit to figure out the proper locations of all. Go down the road to HP. Maybe they are hiring. Some of these pseudo-problem solving questions like this are bunk. I was once asked why sewer covers are round and not square. I gave the correct answer without even hesitation and the interviewer seemed put off that I knew the answer. I didn't get the job but, in hindsight, no great loss. I prefer the questions (like the basketballs one from google) where you won't be able to give an accurate numerical answer but by explaining HOW you would go about solving the problem is all you need to do and MAYBE shows your aptitude for problem solving. Smell the box you opened. Step 1: Order the boxes by weight. Either apples weigh more than oranges, or oranges weigh more than apples. The mixed box will always be in the middle. Step 2: Open the first box, take out the fruit and look at it. Step 3: If the fruit is an apple, deduce that the middle is apple and oranges and that the third box is oranges. If the fruit is an orange, then deduce that the last is the box with the apples. Show More Responses Donna is the only one with any common sense. The problem with corporate America, is that it's run by a bunch of Bozos who over complicate things and have a narrow to zero vision on how to solve even the simplest problems. I can imagine that most of you would get a committee, have long meetings where you talk about 'think out of the box', and 'at the end of the day' nonsense. This is an interesting logic question, but I would not want to buy fruit from a company who knew they had a problem and then sampled one out of three boxes to resolve the issue. There are other correct answers posted. I'll just make a comment: "The boxes have been incorrectly labeled such that no label identifies the actual contents of the box it labels." Nothing in the above statement says the labels are limited to oranges/pears, only that they do not identify the contents. They could say 'nuts', 'bolts', etc. Technically, all answers should be prefaced with: assume that the labels say 'oranges', 'pears', and 'orange/pears'. Ok, the problem does not make sense and is unsolvable if the labels say 'x', 'y', 'z', but someone with (likely with a math proof back ground) may appreciate attention to detail. Q: Why do posters denigrate the interview questions? The questions, however stupid they may be, are a opportunity to show you can build an answer. Even if you pursue an invalid train of thought in the interview, it's a thought. It's what they want to see and what will help you get a job offer. Note: I also would not assume that the questions asked are a reflection on the company, department, or team as a whole. It may just be the interviewer that has chosen poorly. So to say "I don't want to work for company X because they asked me a stupid interview question" is pretty closed minded. To even think I don't want to work with that interviewer just based on questions asked seems extreme. rightly pointed out by Sagmi ... this question was put forward to me at Huawei Technologies and that was the answer I gave So the question was asked at an interview for Apple: Label ALL the boxes apple and charge a ridiculous price for them! Just label all of them "Fruit." Put another way, it is not possible to tell since we don't know how the boxes are mis-labled. What if the Apple box was labeled Oranges and both the other boxes were labeled Apples and no box was labled Apples and Oranges? You might have assumed there are three different labels when their might have only been two different labels. Always pict a piece of fruit from the box labelled Apple&Orange. As we know that this label is wrong, there are two possibilities: If it is apple, then wo know that this box should be labelled Apple, so we switch Apple label with the label Apple&Orange. Then Apple label is correct. We also know that the Orange label is incorrect, so we then switch Orange label and Apple&Orange label. if it is orange, then we know that this box should be labelled Orange, so we switch Orange label with label Apple&Orange. Then Orange label is correct. The same as above, we know that the Apple label is incorrect, so we switch Apple label and Apple&Orange label. If all boxes are labeled incorrectly and u pick a orange out of a box that's labeled apple/oranges change the name to oranges then change the box labeled oranges to apples and the the box labeled apples to apple and oranges... If you pick a apple out of a box labeled apples and oranges change the name to apples and then change the box labeled apples to oranges and the last to apples and oranges... If u pick a apple out of a box labeled oranges change it to apples and oranges then the box labeled oranges to apples and the box labeled apples to oranges...if you pic a orange out of a box labeled apples change it to apples and oranges and the box labeled oranges to apples and the last to apples and oranges... See the pattern? I think there is a big box and it contain two small boxes and all the labels are incorrect so big one contain two boxes that makes it carrys both orange and apples and in that thwo boxes having orange and apple respectively so if we open any box we can label it correctly Show More Responses To see the java source code of puzzle, visit: https://github.com/SanjayMadnani/com.opteamix.microthon code is taking the input by console only. You can fork or clone the repository and proceed further. You can also rise bug if you find any. Run BasketPuzzleGameTest.java class as a Junit test case to start game. if it known already that boxes labeled incorrectly, I would give it back to those who did label them and ask to fix this confusion. it is impossible to tell by opening only one box, so you have to open one more box. As mentioned already, if you start with the A+O bucket, you can solve the puzzle by pulling only one fruit, Bucket: A+O Found: A Bucket A+O > A | A+O, but since A+O label is incorrect, then it must be A Bucket O > since A is taken, the new label must be O | A+O, but since O is incorrect, it must be A+O Bucket A > since A and A+O are taken, it must be O Bucket: A+O Found: O Bucket A+O > O | A+O, but since A+O label is incorrect, then it must be O Bucket A > since O is taken, the new label must be A | A+O, but since A is incorrect it must be A+O Bucket O > since O and A+O are taken, it must be A If you are lucky, you might solve it with just one fruit even if you start with other buckets, Bucket: A Found: A Bucket: A > A | A+O, but since the A label is incorrect, it must be A+O Bucket: O > A | O, but since the O label is incorrect, it must be A Bucket: A+O > since A+O and A are taken, it must be O Bucket: O Found: O Bucket: O > O | A+O, but since the O label is incorrect, it must be A+O Bucket: A > A | O, but since the A label is incorrect, it must be O Bucket: A+O > since A+O and O are taken, it must be A If you start with the A bucket and pull an O or if you start with the O bucket and pull and A, then you are SOL and you need to pull out more fruits to figure it out. 1. Open one box and check its contents. 2. Remove the current label and apply the correct one (by removing it from one of the other boxes) 3. Since all boxes have been labeled incorrectly, switch labels between the other 2 boxes. And Voila you have all the boxes labelled correctly :) In requirement already specify that all three box labels are not correct. A+O A O Step1: First Pick an item A+O Box. If you get an Apple then it is a Apple Box. swap the label . A A+O O AS we already know in the box that label with Orange, does not contain Orange because of wrong label. So It must contain A+O. Just Swap the label A O A+O OK, all 3 boxes are incorrectly labeled. Open the one that says apples and oranges. Whatever is in there is what it is (since it cannot be apples AND oranges). Now, if there was an orange in there, apples must be in the orange box (since they cannot be in the apples box), and apples and oranges in the apples box (due to process of elimination). Get it? I guess questions like these will appear easy if you put them on paper, it is the possible combinations that become relevant, one way to approach is.. One of the key factor is all boxes are labelled incorrectly, this gives rise to only (2) combinations right To label for 1st box incorrectly you will have (2) options, once you label it then you have only choice to label the other two boxes incorrectly so 2 x 1 = 2 combinations possible i.e. Incorrect lablling options { Boxes_with_Oranges, Boxes_with_Apples, Boxes_with_Apples&Oranges } = { A, AO, O} or {AO, O, A} 2. To know for sure the contents of the boxes, you need to pick the box with either Apples or Oranges and avoid box with Apples and Oranges. So from the (2) combinations you could pick a fruit from Box_labeled AO (this will contain either Oranges or Apples) So, if you get a Orange, it means that combination is{AO, O, A} , so that means Box_with_Label_O has Apples, Box_with_Label_A has Apples and Oranges Box_with_label_AO has Oranges or else if you get a Apple that combination is {A, AO, O}. Box_with_Label_AO has Apples, Box_with_Label_O has Apples and Oranges Box_with_label_A has Oranges Then you can correctly label all the 3 boxes. First answer in this post is correct, as its said all boxes doesnt reflect correct items in it, If an apple is picked from a box , then it can be from either A/O box or A box, if the box is names A/O the, the label of the box has to be changed to A, then other two box labels to be accordingly. It is interesting that in 6 years people keep overthinking this. The answer is in the question and the criteria are that the boxes are immediately labeled and they are labeled correctly. ANSWER: FRUIT FYI you don't even need to open one box. Show More Responses Your choice going to be (( 2 apple 1 orange)) or (( 2orange 1apple )) . It can be recognize only one box (x) . U have to chose again until u get another formula then u will named easly . Step 1: Open a box labeled ‘Apples and Oranges’. We know that this box does not contain ‘Mixture’ for sure. If this fruit is an apple, then label this box as ‘Apple’. Step 2: (Very important) If we look at the box labeled as ‘Oranges’, we know that since the label is incorrect, this box either has only apples in it or has Mixture. Since we already know which box contains only apples, we know that the box labeled as ‘Oranges’ contains ‘Mixture’. So label it as ‘Mixture’. Step 3: (Very easy) The 3rd box will be labeled as ‘Oranges’. When you put your hands on the box to pick the fruits by touching every fruits you can feel whether all are apple or oranges or both and just pick one to see.So it is not necessary to pick one fruit and see whether it is orange or apple also it is not said in question that you can touch and feel all the fruits inside the boxes without taking it out .and then you can fix the label correctly on the boxes. Absurd, no logic km I will took a pen and stab the apple. and then ? apple-pen. Smelling the box and writing the correct label on each. :) Just label the boxes correctly. I got this question in a video interview. He verbally read out the question, but either left out the mislabeled part or I not heard it correctly. Without that piece of information, it was impossible to answer the question. Didn't get the job. |

### Software Engineer at Google was asked...

Given the list of points of the skyline of a city in order (from East to West) Find the maximal rectangle contained in this skyline. I was asked to write the code. I managed to find the algorithm but was not sufficient. 21 Answersjust google for skyline. @Interview Candidate Have you found any particularly clear explanations? I'm working on a solution - I'll post it later tonight - but it'd be nice to confirm my understanding. Thanks. Clear? Well, to be more exact: Given N points on 2-plane: (x1, y1), (x2,y2), ... , (xN, yN) s.t. x1=0, xN= 0 and for all 10,yM > 0. Find the maximum value of the area of a rectangle inside the region bounded by these points and the x-axis. You can assume that the rectangle's edges are going to be parallel to the x,y axes. Show More Responses Here's an O(n^3) dynamic programming solution. It's pretty ugly, but it seems to work. More elegant/efficient solutions would be greatly appreciated. -------------------------------------------- import java.util.ArrayList; public class Skyline { private int maxX; private int maxY; public String[] findLargestRectangle(String[] points){ // First translate the points into a boolean matrix. // The cell [i][j] in the matrix should be true if // it appears under the skyline and false otherwise. boolean[][] cells = this.pointsToMatrix(points); System.out.println("Matrix is: "); this.printBooleanMatrix(cells,maxX,maxY); // Now we calculate rectangle size. Define // s[i][j] to be the size of the rectangle // with (i,j) at its upper left corner. int[][] s = new int[maxX+1][maxY+1]; int maxSizeX = 0; int maxSizeY = 0; int maxSize = 0; for (int i = 0; i maxSize){ maxSize = max; maxSizeX = i; maxSizeY = j; } } } System.out.println("Size matrix is: "); this.printIntegerMatrix(s,maxX,maxY); System.out.println("Maximum rectangle has size " + maxSize + "."); System.out.println("Its upper left corner is (" + maxSizeX + "," + maxSizeY + ")."); return null; } // Points to be given as "1 0", "2 5", etc. private boolean[][] pointsToMatrix(String[] points){ ArrayList list = new ArrayList(); // The maximum x-value will be that of the last point. maxX = Integer.parseInt(points[points.length-1].split(" ")[0]); maxY = 0; for (int i = 0; i maxY) maxY = y; list.add(new Point(x,y)); // Assume there are no duplicates } System.out.println("Points are: " + list); System.out.println(" maxX = " + maxX + ", maxY = " + maxY); boolean[][] m = new boolean[maxX+1][maxY+1]; int prevX = 0; int prevY = -1; for (int k = 0; k 0){ m[x][j] = true; j--; } } else { // Look left. If the cell to the left is false, then this cell // is true (start of rectangle top). The value of the cells below // is also true. // // If the cell to the left is true and the previous point had the same value // of x, then this cell represents the start of another rectangle. Set its value // to true and the values of the cells below to true. if ((! m[x-1][y]) || (prevX == x)){ m[x][y] = true; while (j > 0){ m[x][j] = true; j--; } // If the previous value of y is the same as this value of y and the // difference between x values is greater than 1, fill in the cells // in between. if (prevY == y && (x - prevX > 1)){ int i = x-1; while (i > prevX){ j = y; while (j > 0){ m[i][j] = true; j--; } i--; } } } // If the cell to the left is true and the previous point had a different // value of x, then this cell is false (end of rectangle top). Set it and // the cells below it to false. This cell may be inside another rectangle, // in which case its value will be reset to true. else { m[x][y] = false; while (j > 0){ m[x][j] = false; j--; } } } prevX = x; prevY = y; //System.out.println("After " + p + ", matrix is: "); //this.printBooleanMatrix(m,maxX,maxY); } return m; } (cont'd) private void printIntegerMatrix(int[][] b, int maxX, int maxY){ for (int j = maxY; j >=0; j--){ for (int i = 0; i =0; j--){ for (int i = 0; i <= maxX; i++){ System.out.print(String.format("%8b",b[i][j])); } System.out.println(); } } class Point implements Comparable{ int x; int y; Point(int x, int y){ this.x = x; this.y = y; } public int compareTo(Object o){ Point p = (Point) o; return (this.x != p.x ? this.x - p.x : this.y - p.y); } public boolean equals(Object o){ if (! (o instanceof Point)) return false; Point p = (Point) o; return (this.x == p.x && this.y == p.y); } public int hashCode(){ return x*37 + y*19 + x*11 + y*7; } public String toString(){ return "(" + x + "," + y + ")"; } } /** * @param args */ public static void main(String[] args) { String[] points = {"0 2","1 2","1 1","2 1","2 3","3 3","3 0"}; Skyline s = new Skyline(); s.findLargestRectangle(points); System.out.println(); String[] points1 = {"0 1","1 1","1 4","2 4","2 3","3 3","3 6","4 6","5 3","7 3","7 7","8 7"}; s = new Skyline(); s.findLargestRectangle(points1); } } Output: Points are: [(0,2), (1,2), (1,1), (2,1), (2,3), (3,3), (3,0)] maxX = 3, maxY = 3 Matrix is: false false true false true false true false true true true false false false false false Size matrix is: 0 0 3 0 2 0 2 0 3 2 1 0 0 0 0 0 Maximum rectangle has size 3. Its upper left corner is (0,1). Points are: [(0,1), (1,1), (1,4), (2,4), (2,3), (3,3), (3,6), (4,6), (5,3), (7,3), (7,7), (8,7)] maxX = 8, maxY = 7 Matrix is: false false false false false false false true false false false false true false false false true false false false false true false false false true false false true false true false false false true false false true true true false true true true false false true true true false true true true false true true true true false true true true false false false false false false false false false false Size matrix is: 0 0 0 0 0 0 0 7 0 0 0 0 6 0 0 0 6 0 0 0 0 5 0 0 0 5 0 0 4 0 4 0 0 0 4 0 0 9 6 3 0 9 6 3 0 0 6 4 2 0 6 4 2 0 4 3 2 1 0 3 2 1 0 0 0 0 0 0 0 0 0 0 Maximum rectangle has size 9. Its upper left corner is (1,3). Found bug: public *int* findLargestRectangle(String[] points){ ... *return maxSize*; } (obviously) I don't mean to discourage you but there is no way that you can get away with an answer that works in O(N^3) for this question and even if the interviewer was happy with this performance, it would be impossible for you to write this kind of code given a reasonable time limit for the interview (which is around 45 minutes, and that's the entire interview which includes introduction and possibly some other easy questions) You can solve this in O(N). I don't have time to write a solution right now, I strongly recommend looking at Topcoder's skyline problems and contestants' solutions for an insight. best of luck to you For others' reference: http://www.topcoder.com/stat?c=problem_statement&pm=7473&rd=10661 http://www.topcoder.com/tc?module=Static&d1=match_editorials&d2=srm337 https://www.spoj.pl/problems/HISTOGRA/ http://www.informatik.uni-ulm.de/acm/Locals/2003/html/judge.html Here's a much nicer O(n^2) solution. As Anonymous said, there's an O(n) solution, too. /* * In this input array a, the ith building * has height a[i], which means its lower left * corner is at (i,0) and its upper right corner * is at (i+1,a[i]). All buildings have width 1. * The total width of the skyline is n. */ public long getMax(long[] a){ int n = a.length; int[] leftWidths = new int[n]; int[] rightWidths = new int[n]; // For each building, check how many units width to the left of the // top block of this building we can move before we are no longer // in a building. for (int i = 0; i = 0) && a[x] >= a[i]) { x--; count++; } leftWidths[i] = count; } // For each building, check how many units width to the right of the // top block of this building we can move before we are no longer // in a building. for (int i = 0; i = a[i]) { x++; count++; } rightWidths[i] = count; } // Now find the maximum rectangle. The value of the ith building's // rectangle is its horizontal width times its height, or // (1 + leftWidths[i] + rightWidths[i]) * a[i]. long maxArea = 1; for (int i = 0; i maxArea) { maxArea = nextArea; } } return maxArea; } O(n) solution: /* * Basic idea: use a stack to store the buildings. Look at * the buildings in left-to-right order (west to east). If a * building is taller than the building on the top of the stack * (the tallest building to its left), push it onto * the stack. If a building is equal in height to the building on the * top, skip it. If a building is shorter than the building on the top, * it is not part of the maximum rectangle that is topped by the tallest * building to its left. Pop that tallest building, calculate its area and * compare it to the current main area, then repeat the comparison * procedure with the new tallest building. * * Along the way, track the number of buildings to the left and right of a * given building that would participate in that building's maximum * rectangle. The number to the left is equal to the number of buildings * that are popped off the stack before this building is pushed - that is * the number of buildings to the left of this building that are taller. * We do not need to worry about the buildings that are equal in height * since they are discarded (they are accounted for in the topBuilding's * rightWidth count). * * The number of buildings to the right of this building that participate * in this building's maximum rectangle is equal to the number of buildings * that are discarded because they are equal to this building's height * plus the number of buildings that are pushed onto the stack because they * are taller than this building while this building is on the top of the * stack. * * In this input array a, the ith building has height a[i], which means * its lower left corner is at (i,0) and its upper right corner is at * (i+1,a[i]). All buildings have width 1. The total width of the skyline * is n. */ public long getMax_useStack(long[] a){ Stack stack = new Stack(); int n = a.length; long maxArea = 1; // Process the buildings in left-to-right order. for (int i= 0; i < n; i++){ Building nextBuilding = new Building(a[i]); // Keep track of the number of buildings that we pop before we // push nextBuilding. That number will be equal to the number // of buildings to the immediate left of nextBuilding that are // taller in size. int popCount = 0; // If the stack is empty, push the next building onto the stack. // There are no buildings to its left, so we do not need to // update nextBuilding.leftWidth. if (stack.empty()) { stack.push(nextBuilding); continue; } (cont'd) // Otherwise, compare the new building's height to the building on // the top of the stack until the new building is either // discarded (if it is equal in size) or pushed (if it is taller). while (! stack.empty()){ Building topBuilding = stack.peek(); long heightDiff= nextBuilding.height - topBuilding.height; // If the new building is equal in height, skip it. Increment // the rightWidths count of topBuilding as its largest // rectangle goes through the new building. if (heightDiff == 0) { topBuilding.rightWidth++; break; } // If the new building is greater in height, push it onto the // stack. The number of buildings to the immediate left of it // that are taller is equal to the number of buildings that // were popped before this point, its popCount. Set its // leftWidth equal to its popCount. Increment the rightWidths // count of the top building as its largest rectangle goes // through the new building. if (heightDiff > 0) { nextBuilding.leftWidth = popCount; topBuilding.rightWidth++; stack.push(nextBuilding); break; } // If the new building is less in height, update the maximum area // with regards to the element at the top of the stack. long topArea = topBuilding.getArea(); if (topArea > maxArea){ maxArea = topArea; } // Then discard the top element and repeat the comparison // procedure with the current next building. stack.pop(); popCount++; } } // If all buildings have been processed and the stack is not yet empty, // finish the remaining subproblems by updating the maximum area // with regards to the building at the top of the stack. while (! stack.empty()){ Building topBuilding = stack.pop(); long topArea = topBuilding.getArea(); if (topArea > maxArea){ maxArea = topArea; } } return maxArea; } class Building { long height; int leftWidth; int rightWidth; Building(long y){ this.height = y; leftWidth = 0; rightWidth = 0; } long getArea(){ return height * (1 + leftWidth + rightWidth); } } ellemeno, Nice solution/explanation. However, maxArea should be initialized to 0. @logimech Right. :) Thanks. Show More Responses ellemeno: The program has bug. We need to store the building's x cordinate. And when the nextbuilding is shorter than the top one, we need to update each popped building's rightwidth as topbuilding.x - nextbuilding.x Isnt this the convex hull problem? The O(n) algorithm does not work. I have another solution for it, however. Suppose we have the array of heights a of length n. We are looking for the smallest building index (let's say x and divide the problem in 2: the left part ( between 0 and i - 1) and the right part (between i + 1 and n - 1). Then we compute the area that can be build using the building indexed with i: a[i] * (n - 1 - 0 + 1) = n * a[i]. We compare this area with the ones obtained from the left and from the right part and return the result. Therefore, the complexity will be: N * time to find the index of the smallest building in a range - which is RMQ problem that can be solved in O(logn) and O(n) preporcessing time (and O(N) space). Therefore, the complexity is O( N log N). It works for 3 6 5 6 2 4: the solution is 3 * 5 = 15. The problem is known as the biggest rectangle below a histogram. An old question with my answer. using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace SkylineTest { class Program { static void Main(string[] args) { Point[] pointList = new Point[10]{new Point(0, 1) ,new Point(1,1) ,new Point(1,2) ,new Point(2,1) ,new Point(2,2) ,new Point(2,0) ,new Point(4,0) ,new Point(4,6) ,new Point(5,6) ,new Point(5,0)}; Skyline sl = new Skyline(pointList); Point maxPoint = sl.CalculateMaxRanctangle(); Console.WriteLine("MaxPoint:" + maxPoint.X + " " + maxPoint.Y); Console.WriteLine("MaxArea:" + sl.MaxArea); Console.ReadLine(); } } public struct Point { private int _x; private int _y; public int X { get { return _x; } } public int Y { get { return _y; } } public Point(int x, int y) { _x = x; _y = y; } } public class Skyline { private Point[] _pointList; private int _maxArea=0; private Point MaxRactanglePoint; public int MaxArea { get { return _maxArea; } } public Skyline(Point[] pointList) { _pointList = pointList; } public Point CalculateMaxRanctangle() { Point maxPoint = new Point(0, 0); for (int index = 0; index 2) { for (int i = index - 2; i >= 0; i -= 2) { if (_pointList[i].Y >= _pointList[index].Y) areaForPoint += (_pointList[i+2].X - _pointList[i].X) * _pointList[index].Y; else break; } } if(index= _pointList[index].Y) areaForPoint += (_pointList[i].X - _pointList[i-2].X) * _pointList[index].Y; else break; } if (areaForPoint > _maxArea) { _maxArea = areaForPoint; maxPoint = _pointList[index]; } index += 2; } } return maxPoint; } } } i dont know if i understand the assignement correctly, but shouldnt ne sufficient to iterate all coorfinates and find min x, min y, max x, max and then calculate the area given by this rectangle? Hi Guys, there is O(N) solution using STACK. http://www.geeksforgeeks.org/largest-rectangle-under-histogram/ Psuedocode //Iterate through points one time where a point further east is greater than a point farther west, a point higher up is greater than a point lower down. //if x greatestX, greatestX = x //if y greatestY, greatestY = y //Equate (greatestX - smallestX) by (greatestY - smallestY) For some reason that left out like half of my psuedocode |

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