Interview questions in San Jose, CA
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Interview Questions in San Jose
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. 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. One or more comments have been removed. |
Data Scientist Intern at LinkedIn was asked...
Find the second largest element in a Binary Search Tree 16 Answersfind the right most element. If this is a right node with no children, return its parent. if this is not, return the largest element of its left child. One addition is the situation where the tree has no right branch (root is largest). In this special case, it does not have a parent. So it's better to keep track of parent and current pointers, if different, the original method by the candidate works well, if the same (which means the root situation), find the largest of its left branch. if (root == null || (!root.hasRightChild() ) { return null;} else return findSecondGreatest(root, root.getValue()); value findSecondGreatest(Node curr, value oldValue) { if(curr.hasRightChild()) { return (findSecondGreatest( curr.getRightChild(), curr.value)); } else return oldValue; } Show More Responses The above answer is also wrong; Node findSceondLargest(Node root) { // If tree is null or is single node only, return null (no second largest) if (root==null || (root.left==null && root.right==null)) return null; Node parent = null, child = root; // find the right most child while (child.right!=null) { parent = child; child = child.right; } // if the right most child has no left child, then it's parent is second largest if (child.left==null) return parent; // otherwise, return left child's rightmost child as second largest child = child.left; while (child.right!=null) child = child.right; return child; } Soln by "mindpower" works. Thank you. I am trying to solve a similar problem Find the 2nd nearest high(in in-order traversal) value for a given node Eg: Given nums: 12 7 14 3, construct a BST. If the given value is: 7 then we should return 14 (in the sort order: 3, 7, 12, 14) if the given value is: 3 then we should return 12 (in the sort order: 3, 7, 12, 14) Generic solution in C# for any k. Notice that this example can be easily changed to find the k-th smallest node by doing a depth-first recursion on root.Left first, and then a tail recursion on root.Right. public Node GetKthLargest(int k) { return GetKthLargest(ref k, this.Root); } Node GetKthLargest(ref int k, Node root) { if (root == null || k < 1) return null; var node = GetKthLargest(ref k, root.Right); if (node != null) return node; if (--k == 0) return root; return GetKthLargest(ref k, root.Left); } recursion is not needed. SecondLargest(Node root, Node secondLarge) { if(root.right==null) return root.left; Node secondLargest = root; while(secondLargest.right.right==null) secondLargest=secondLargest.right; return secondLargest; } int getmax(node *root) { if(root->right == NULL) { return root->d; } return getmax(root->right); } int secondmax(node *root) { if(root == NULL) { return -1; } if(root->right == NULL && root->left != NULL) { return getmax(root->left); } if(root->right != NULL) { if(root->right->right == NULL && root->right->left == NULL) { return root->d; } } return secondmax(root->right); } In-order traverse the tree. The second last element in the array in the answer. In Python: def find_second_largest_bst_element(root, parent=None): if parent is None: # BST root if root.right is None: # no right subtree if root.left is not None: # if a left subtree exists... return root.left else: # root is the only element of the BST return False else: if root.right is None: # right-most element if root.left is not None: # left subtree exists return root.left else: # leaf return parent else: # check right subtree find_second_largest_bst_element(root.right, root) find_second_largest_bst_element(root) For kth smallest, descend the left subtree first. class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right def findKthLargest(root, k): global count if root is None: return findKthLargest(root.right, k) count += 1 if count == k: print root.value return findKthLargest(root.left, k) count = 0 r = Node(10, Node(5, Node(2), Node(7)), Node(30, Node(22), Node(32))) findKthLargest(r, 3) // solution in java // main routine Node findSecondMax(Node root) { if(root == null || (root.left == null && root.right == null) return null; else { Node max = findMax(root); return (max.parent == null) ? findMax(max.left) : max.parent; } } //helper routine, recursive implementation.... can also be done non-recursively Node findMax(Node root) { return (root.right == null) ? root : findMax(root.right); } Show More Responses Find the largest number in the binary tree and delete it. And again find the largest number. Short and fast. Reverse in-order traversal of the BST, keeping a count of # of visited nodes. This methods works great to return the kth largest element in a BST. mindpower's solution looks right One or more comments have been removed. |
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; } One or more comments have been removed. |
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 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. One or more comments have been removed. |
Data Analyst at Google was asked...
If you have 10 bags of marbles with 10 marbles each and one bag has marbles that weigh differently than the others, how would you figure it out from one weighing 7 Answershmm, drop all the bags a the same height spread out and see which bag make the biggest mark on the ground. Add bags one at a time to scale. Should be the same increment of weight added until you add the bag that adds a weight value different from all the others. hint: use the number of stones to code for each bag. Show More Responses Assume all marbles are from 10g and the heavier one is 11g Take 1 marble from bag1 ,2 from bag2, 3 from bag3, 4 from bag4, 5 from bag5, 6 from bag6, 7 from bag7, 8 from bag8, 9 from bag9,10 from bag 10..and weigh them together Let it be W. So if bag5 contains the heavy marbles The total weight (W) will be 10+20+30+40+55+60+70+80+90+100 = 555. where as if all were of 10g it should have been 550. meaning the bag which is heavy will always be MeasuredWeight - 550 Mathematically if bag X is the one which is heavy, X can be found using one weighing of sample (W) - (N(N+1)/2) Terribly worded question; you never specified that we are given the normal and heavy weights. Without that information, the votes up solution here doesn’t work. Using a series sum [N(N+1)/2], will not work in this case, as we've not been provided with enough specifics (such as "whats the standard weight of the marbles?). In fact any solution dependent on summed weight will have overlapping solution space and fail. Consider if the standard weight is 10 ounce per marble, and bag 1 is the exception at 12 ounce per marble, for a total weight of 552. Now consider if bag 2 is the exception at 11 ounce per marble - also for a total weight of 552. There is no way to distinguish between the two cases. Given the setup, if appears the interviewer did expect the answer to use some variation of summed series, and I suspect the poster has paraphrased the question and missed some key language - and as best I can tell is not solvable as stated. Google will expect you to provide a generalized solution that can be automated, so the "add the bags one at a time", while simple and clever, would probably not be acceptable as a "weigh only once" solution, and if accepted would be a lesser answer as it does not provide a code-able business solution. One or more comments have been removed. |
Product Manager at Google was asked...
You notice that adwords revenue for a certain word has dropped in Italy for the last 30 days. How do you go about determining why that has happened? 4 AnswersThis is a test to see how you think on your feet. Adword Revenue : No. of impressions * Click Thru Rate * Cost Per Click Anyone of the three parameters could decline to have an overall reduction in revenue. No. of Impressions could go down if a. The internet usage has fallen for some socio-culturaal reasons in Italy b. The usage of Google search has reduced because of may be some competitor applicaton launch or some major marketing promotion activities c. Some major technical issues has come up may be in the Google servers which is resulting in higher latency in Google Search applications resulting in reduced usage Click Thru Rate might have gone down 1 Major shift is usage clusters Keywords used have changed resulting in changed search behavior where in people are less prone to click thru. 2 Some technical issues like adds not displayed properly Same major flaw in random add picking might have got introduced 3. Some recent layout change has been there and peope are yet to get accustomed with the changed layout Reduced CPC: 1. People are spending less on Adwords and hence bidding less 2 Due to the keyword change, the new cluster CPC is much lesser. 1. Determine the amount of decrease in month over month percentages, and make sure this isn't a trend. 2. Assuming we've seen similar decreases in conversions and clicks, 30 days is a month's time. Let's say this is in August, when the entire country uses the majority of their average of 42 vacation days per year. That's a factor. 3. Given you've said decreases in revenue and assuming all click and conversion data remained the same month over month, we may look for broken dynamic revenue variable conversion codes on the page source. Was there a site update? Show More Responses One or more comments have been removed. |
Product Design Engineer at Apple was asked...
What are the different ways you can you tell if this part is steel or aluminium. 4 AnswersSimply by using a magnet, Steel has metallic properties, and the magnet will connect. Aluminium will do nothing. Many stainless steel alloys are not magnetic, so if your magnet is attracted to the material you will definitely know it is steel, but if it doesn't you will not know what the material is. Simple methods would be density (feeling the mass of the object), surface finish (color, texture). If coated that may give you the answer i.ie anodized would indicate aluminium. For more information I would go for EDX( Energy-dispersive X-ray spectroscopy) and possible a cross section to look at the grain structure. Show More Responses By far the easiest way is to test for material properties. -density -hardness -modulus of elasticity I would choose hardness. Strike each item with an equal force, which one deforms more? Thats aluminum. You could probably pull this test off with a hammer. The simplest solutions is always the best. |
Product Design Engineer at Apple was asked...
What are 5 different variables of a coil spring you can change to affect spring force? 3 AnswersMaterial, wire diameter, wire cross sectional shape, coil diameter, coil length 5 different variables that can affect the springs force. 1.) Wire length 2.) Coil pitch 3.) Wire cross-sectional shape 4.) Material of the coil 5.) Coil diameter 1. WD 2. OD 3. Active coils 4. Material 5. Free Height |
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 One or more comments have been removed. |