# Internet interview questions

## Interview Questions

### 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))) |

### 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 Above answer is wrong. it has to be something like this. public static int findSecondLargest(Node node) { Node secondLargest = null; Node parent = null; Node child = node; if (node!=null && (node.hasLeftChild()||node.hasRightChild())) { if (node.hasRightChild()) { while (child.hasRightChild()) { parent = child; child = child.rightChild(); } secondLargest = parent; } else if (node.hasLeftChild()) { child = node.leftChild(); while (child.hasRightChild()) { child = child.rightChild(); } secondLargest = child; } } return secondLargest; } 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 |

### 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? 6 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 |

Describe and code an algorithm that returns the first duplicate character in a string? 7 AnswersSimple Python example. Not sure it's most efficient. def findDup(str): match=[] i=1 while (i<len(str) and len(match)==0): for j in range(i): if str[j]==str[i]: match=[j,i] i+=1 return match if __name__ == '__main__': print findDup('asdf') print findDup('asdfasdf') pass first clarify if it is ASCII or UNICODE string For ASCII, create BOOL checkArray [128] = {false}; walk the string and update the index of checkArray based of the character. for (int index=0;index< strlen(str); index++) { if (checkArray[str[index]] == true) { printf (str[index]); return; } else { checkArray[str[index]] = true; } } public class FirstDupCharacter { public static void main(String[] args) { System.out.println(findDupCharacter("abcdefghiaklmno")); } private static Character findDupCharacter(final String input) { final Set set = new HashSet(); Character dup = null; for (int i = 0; i < input.length(); i++) { if (set.contains(input.charAt(i))) { dup = input.charAt(i); break; } else { set.add(input.charAt(i)); } } return dup; } } Show More Responses String samp = "Testing"; samp = samp.toLowerCase(); char chararr[] = samp.toCharArray(); int size = chararr.length; char repeat = ' '; for (int i=0;i<size && repeat==' ';i++) { for (int j=i+1;j<size && repeat==' '; j++) { if (chararr[i]==chararr[j]) { repeat = chararr[i]; } } } System.out.println("First Repeating character :: "+repeat); for (int i=0;i<size;i++) { if (samp.indexOf(chararr[i]) != samp.lastIndexOf(chararr[i])) { System.out.println("First repeating char ::"+chararr[i]); break; } } public static in findDuplicateChar(String s) { if (s == null) return -1; char[] characters = s.toCharArray(); Map charsMap = HashMap(); for ( int index = 0; index < characters.length; index++ ) { // insert the character into the map. // returns null for a new entry // returns the index if it previously if it existed Integer initialOccurence = charsMap.put(characters[index], index); if ( initialOccurence != null) { return initialOccurance; } //there where no characters that where duplicates return -1; } } Another python solution: def findFirstNonRepeatedCharInOneIteration(str1): for i,j in enumerate(str1): if j in str1[:i] or j in str1[i+1:]: print "First non-repeated character is "+ j break str1 = "abcdefglhjkkjokylf" findFirstNonRepeatedCharInOneIteration(str1) |

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

What sort would you use if you required tight max time bounds and wanted highly regular performance. 6 AnswersVector sort. Guaranteed to be O(n log n) performance. No better, no worse. That is so say, a "Balanced Tree Sort" is guaranteed to be O(n log n) always. Show More Responses Merge sort and heapsort are always guaranteed to be n*log(n). Quicksort is usually faster on the average but can be as bad as O(n^2), although with very low probability. Heapsort also does it sorting in-place, without needing an extra buffer, like mergesort. Lastly, heapsort is much easier to implement and understand than balancing trees mentioned by earlier posts. for something like this you generally want bubble sort or insertion sort. It's not about being fast it's about being consistent. Make it do exactly the same thing every time. Use a sorting network. There's some precomputation time, but runtime will be very consistent (the only variability is branch prediction performance) |

### 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) |

### 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 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) |

### IT Manager at Hightail was asked...

Why are point to point VPNs not exactly the best way to connect LANS 2 AnswersThe rekeying that occurs plays havoc with monitoring software. The rekey interval is usually 1 or 8 hours, by default. It can be made longer if desired. The biggest downsides I see of VPNs are 1) No firm SLA - VPNs are dependent on the Internet, and thus prone to any performance issues or outages. 2) Limited Scalabilty - As the network expands or changes, all the tunnels must be manually updated (unless you're running a dynamic routing protocol across them). 3) Limited features - For example, it's impossible to bridge the same subnet across a VPN. 4) Complexity - IPSec has lots of options, and if both sides don't match exactly, the tunnel will have problems. This is a big headache if you don't control the equipment on both sides of the tunnel. |

What is the most important part of the sales cycle to you? Qualification, Presentation, Negotiation or Close? 1 AnswerFor me, qualification was my answer. I don't want to present, negotiate or try to close an unqualified prospect. |