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Apr 24, 2011
 Print a singly-linked list backwards, in constant space and linear time.10 AnswersOne can create a function that takes a node as an argument and checks whether the next of the passed node is NULL or not.In case it is not NULL,the same function is again called for the NEXT node.if the Next of the passed node is NULL,the function prints the value of the node and returns. void f1(node* p) { if(p->next!= NULL) { f1(p->next) } else { print ("%d",p->value); } }Arpit, that's not constant space, that's linear (stack) space, since you will have as many function calls waiting to be returned on the stack as there are nodes. The trick is to reverse the list first (constant space, linear time when done iteratively or tail-recursively) then print it in order (against constant space, linear time).void f1(LinkedListNode lln) { if(lln.next != null) f1(lln.next); System.out.print(lln.value); }Show More ResponsesBobby, that is not constant space because it uses O(N) stack space. There are obvoius O(N^2)-time O(1)-space algorithms, and obvious O(N) time O(N) space algorithms. This is my best guess. Assuming you have exclusive access to the list, you can reverse it, walk it, and then reverse it again. Something like this: #include #include struct node { int value; struct node * next; }; void print_backwards( node * head ) { node * prev = NULL; node * cur = head; node * next; while( cur ) { next = cur->next; cur->next = prev; prev = cur; cur = next; } cur = prev; prev = NULL; while( cur ) { printf( "%d\n", cur->value ); next = cur->next; cur->next = prev; prev = cur; cur = next; } assert( prev == head ); } main() { node a, b, c; a.value = 1; a.next = &b; b.value = 2; b.next = &c; c.value = 3; c.next = NULL; print_backwards( &a ); }Reverse the list, print, then reverse it back. 3 O(n) operations = O(n). O(1) space used.Reverse the list, print, then reverse it back. 3 O(n) operations = O(n). O(1) space used.void BWDisplayLinkedList(node* pHead) { if(!pHead) return; BWDisplayLinkedList(pHead->next); cout data "; }Does this work well? Reverse a list, print it, and then reverse it again. struct node *reverse(struct node *oldlist) { struct node *newlist = NULL; while(oldlist!=NULL) { struct node *temp = oldlist; oldlist=oldlist->next; temp->next=newlist; newlist=temp; } return newlist; } void display(struct node **q) { struct node *temp; temp = *q; if(*q==NULL) { printf("List is empty\n\n"); } else { while(temp!=NULL) { printf("%d=>",temp->data); temp=temp->next; } printf("||\n\n"); } } //p is our list p = reverse(p); display(&p); p = reverse(p);Thanks to recursion :) void print_backward(node* n) { if(n == NULL) return; print_backward(n->nxt); cout val << endl; }Yes recursion does the job in linear and constant time. :-)

Apr 24, 2011

Jan 27, 2012
 Describe a routine which returns the set of integers in {1..100} divisible without remainder by 3 but not by 9.12 AnswersI'm assuming the question wants us to find integers that are divisible by 3 but not by 9. This can be easily obtained using a mod function inside the following if statement: if(number % 3 == 0 && number % 9 != 0) Here is a short program I wrote in c++ to show how to solve this problem. Instead of returning the set of integer, I just printed them out on the screen: #include #include using namespace std; int main(int argc, char** argv) { int i = 0; vector list; vector::iterator it; for(i = 1; i <= 100; i++) { if(i%3 == 0 && i%9 != 0) { list.push_back(i); } } for(it = list.begin(); it != list.end(); it++) { cout << *it << endl; } return 0; } If I missed anything, please let me know. Happy coding and problem solving!That'll certainly work, Tyler, but the OP indicated he was interviewing for a Ruby On Rails - not C++ - gig.put those integers into an array, pick every third element, out of which discard every third element.Show More Responsespython [x for x in range(0,100) if x % 3 == 0 and x % 9 != 0]1) start from number = 3 Loop while(number <= 100) 2) display number 3) number = number+3, display number 4) number = number+6 Loop(1..100).map { |i| (i % 3).zero? && !(i % 9).zero? ? i : nil }.compact(1..100).select { |x| x%3 == 0 && x%9 != 0matt has the best answerA variation on Matt's answer: (1..100).select { |n| n % 3 == 0 }.reject { |n| n % 9 == 0 }The requirement doesn't say if the input has to be a Range. If it doesn't have to be, then we don't need to traverse each element but to simply calculate it. def get_nums_by_3_not_by_9(max) arr = [] x = max.to_i / 3 x.times do |i| next if i % 3 == 0 arr << i * 3 end return arr end(1..100).select do |n| n%3 ==0 and n%9 != 0 end(1..100).to_a.delete_if{|x| !(x%3==0 && x%9>0)} or (1..100).to_a.select{|x| x%3==0 && x%9>0} or (1..100).to_a.map{|x| x%3==0 && x%9>0 ? x : nil}.compact or (1..100).to_a.reject{|x| !(x%3==0 && x%9>0)}

Feb 25, 2012
 Find the second largest element in a Binary Search Tree16 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 ResponsesAbove 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 ResponsesFind 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

Software Engineer at Amazon was asked...

May 12, 2010
 Given a list of integers, some of which may be negative, extract the pair that sums to the largest number.8 AnswersThe naive solution is O(n^2). The trick is to sort the list in O(n lg n) then pick the two largest numbers from the front.you can get the largest two number in O(n), right? sum those two numbers up.Whoops, I wrote the wrong question. Here's what I meant to say: given a list of integers, find a pair that sums to a given value.Show More ResponsesKeep a list of integers, and set it to 1 if they are in within the list. for (int i = 0; i < n; ++i) dp[array[i]] = 1; for (int i = 0; i < n; ++i) if (dp[S - array[i]] && S - array[i] != array[i]) print S - array[i] and array[i] because they sum to S This is the subset problem. This is O( n ), but requires O( S ) space.@bleh: I guess the solution will work if all the elements are +ve, in this case however the elements are negative. So probably we can use hash instead of arrays.if both hashing and the subset solution aren't good enough - 1. Sort the array o(n) or o(nlogn) - pick the one you can sell 2. Have two pointers - one at the end and the other at the beginning 3.a. If the sum is less than S increment the one in the beginning 3.b. If the sum is greater than S decrement the one at the end 3.c If the sum is S - you are done 3.d If the two pointers have met or crossed over - you are doneAmazon really loves dynamic programming eh? I've come across many interview questions with the knapsack and coin-change problems#define SIZE 3 int tab[SIZE]; int sum; int max=-MAXINT; int sovi, sovj; for(i=0; i max) { max=sum+tab[j]; sovi=i; sovj=j; } } } printf("\n i: %d j:%d", sovi, sovj);

Software Development Engineer at Amazon was asked...

Jan 14, 2010
 How would you find the pairs of numbers that added to some specific number in an array. 7 Answersi program in java..so i will talk from the arrayLists perspective. suppose we want to find out a+b = 123; for(int i=0; i2000 records.but below that, sorting and above operation is efficient. you must play with different possibilities.Your answer (using arrayList.indexOf(...)) is worse than sorting. Sorting is O(log n), finding an item in an unsorted array using ArrayList.indexOf is O(n). Given an unsorted array input (e.g. int[] input), sort it using Array.sort(input) ... this is O(log n). Start at input[0] of the sorted array and calculate it's complementary value. Go to the end of the array and iterate backwards until you find the complementary value or less. If it's less repeat for input[1] and iterate backwards from the previous last item ... keep going. This is at worst proportional to n/2, ie O(n).I realize I wasn't totally clear in my first paragraph ... searching for the complementary value of one item is O(n), but you have to the search for (at worst) every item, so your solution is O(n^2).Show More ResponsesDuh - sorting is O(nlog n) ...Using hashing, this can be done in O(N). store the n/2 mod i in the corresponding hash bucket. Now check the each hash bucket and you are done.sort the array (o(n(log(n)) and take two pointers at both the ends say p1 and p2 if p1+ p2 move the pointer p2 by 1 and add to check if (p1+p2) > n -> move the pointer p1 by 1 and add to check if (p1+p2) = n ->print the pair [traversal takes o(n)] finally thus can be done in o(n)I will not waste o(nlogn) in sorting the array. Instead assuming that the sum we looking for is k, i will divide the array into 2 arrays. First array will contain all values which are less than k/2 Second array will contain all values > k/2 This is bcoz in a sum if one number is less than k/2, the other has to be larger. I will iterate over the smaller array of the 2 since they would rarely be equal. For each x in array 1, i will find the k-x in array 2. Complexity will be O(n).

Software Engineer at Amazon was asked...

Dec 24, 2011
 If you have a file containing millions of integers, how would you sort the data in the file using extremely limited resources, such a s 1GB of memory?8 AnswersWhat was the answer? In my opinion, I would answer this way: Sort each x integers (x is the number of integers that the memory can hold) then save it to the a file (n files) . Then for each file maintain a current index initiated with 0. Loop through n file and take the max integer, put it into the result file, and increase the file's current index to one. Continue doing it till the end.An integer is only 4 bytes! I am just saying, a million integers should consume about 4 MB of memory, and easily fit in 1 GB! So "millions" had better be closer to a quarter billion if memory is an issue. OK, to answer the intended question, I would solve it with statistics. Compute bucket ranges by using a random subset of the file, with bucket size chosen such that the expected portion of the entire file will fit within the memory available. Divide the file into these buckets, and sort the buckets individually. Concatenate the sorted buckets to create a single sorted file.To Kurt: You missed one step. Concatenating buckets will not make a sorted list. You have to merge the buckets. However, for merging you do not necessarily have to load the whole bucket into memory.Show More Responsesmillions is not too much , you can build a int array which have millions of cells. The array take only several mega bytes . Initialize to zero for every cell, then traverse the file and keep track of every integer by array , eg . 4789 will make array[4789]++ , this could handle duplicate number . Finally scan the array ...the interviewer is looking for the word "divide and conquer". read the n number of integers and write them in a binary tree. hold the left and right value in a datastructure which keeps track of the file an integer is written to. something like this class NumHolder { int min; int max; String fileName; } Map partInfo = new HashMap() for (int i=0; i 100000 //100000 = sum number that determines how many part files will be created. writeToOtherFileToo writeToPartFile update partInfo with this write }its called 'external sorting'... read more on wiki...divide the file in chunks, sort individuals chunks using quicksort and store in files. Then use merging step of merge sort on those saved files. This would be external sort.use external sorting algorithm similar to merge sort which splits and merges files instead of arrays in memory

Aug 16, 2010
 There are n pots with different # gold coins in them. Two players play a game, where each player can select a pot at either ends. maximize the gold7 AnswersUse dynamic programming to solve the same. Could explain the logic, but again did not find the time to code it up.What does "maximize the gold" mean in this game? If it means to keep the heavier pots for the end of the game and keep the two sides balanced, then I think you need to sort the N pots and redistribute so that the heaviest is in the middle and the two sides are balanced. Since a pot can contain only a limited number of gold, I would use a count sort algorithm for an average time of O(N+k). When re-distributing, I would balance the two sides starting by the heaviest pot, according to the individual sides's weight. For example: 1, 4, 10, 2, 3, 4, 2 Counting sort: 1:. 2:.. 3:. 4:.. 5: 6: 7: 8: 9: 10:. Sorted list: 1, 2, 2, 3, 4, 4, 10 If K is odd, remember the heaviest = 10 Re-distributing by weights: 4, 3, 1 4, 2, 2 Final list: 1, 3, 4, 10, 4, 2, 2To maximize the gold means, each player wants to maximize the amount of gold the player gets by selection of one pot at a time (either leftmost or rightmost). Player 1 selects a pot, followed by player 2, followed by player 1 etc. How does the player select picking Left or right one ? Maximize gold means sum of the gold coins of all the pots selected by a player. I did not follow how sorting helpsShow More ResponsesHow about take differences and get the side that gives you a better gain. For example: 8 7 4 6 Take 6 Then take 7I think the idea is that each person takes turns choosing the currently leftmost or rightmost pot and tries to maximize the sum of all the pots they take. You can't be greedy because of situations like this: 10, 100, 10, 5 If you were greedy, you would take 10 first, and the other person would take 100. If you instead took 5, they would be forced to take one of the 10's, leaving the 100 available for you to take. The recursive idea here is minimax. The maximum I get from X to Y is the maximum of either the gold at X + what's left over when the other player takes from X+1 to Y, or the gold at Y + what's left over when the other player takes from X to Y-1. As an implementation note, we should keep track of the amount both players get over each interval. If the other player gets R gold, with T gold leftover for me, when choosing from the subrange X+1 to Y, then from the range X to Y, I can easily say that if I choose to take the gold from X, I get T + gold[X] gold, leaving R gold for the other person. This makes the code much simpler. Since each case is based only on subcases of a smaller length (e.g. from X to Y, which is length Y - X, only depends on X to Y-1 and X+1 to Y, both of length Y-X-1), we can build up the cases in order of length. C++ DP implementation would look something like this: pair dp[n][n]; for (int i = 0; i L = dp[i + 1][i + len]; pair R = dp[i][i + len - 1]; if (gold[i] + L.second > gold[i + len] + R.second) dp[i][i + len] = make_pair(gold[i] + L.second, L.first); else dp[i][i + len] = make_pair(gold[i + len] + R.second, R.first); } } The amount of gold I get is dp[0][n - 1].firstUse induction: Suppose that I know how to play best with N arbitrary pots. What's my best move with N+2 pots? Trivial cases: N=0 and N=1. With N+2 pots, I can chose L or R. After that my opponent chooses L or R. Then it's my turn again, but this time, with N pots, I know how to play, by induction hypothesis. So what's the best move? It's the one that maximizes the worst case scenario (that is, my opponent playing her best). What's the worst if I pick the left? It's the value of the left pot plus the minimum of two scenarios: 1. the opponent picks L and I do my best with what's left and 2. the opponent picks R and I do my best with what's left. I do the same calculation assuming that I take the right first. The best of the two worst scenarios maximizes my goal. Here's the code in C++: enum Move ( L, R, E }; // left, right, end of game Move BestMove(int const * left, int const * right, int & minGain) { if (left == right) // there's nothing left in between { minGain = 0; return E; } if (right - left == 1) // just one left { minGain = *left; return L; // arbitrary choice } // I take left, she takes right int xLL; BestMove(left + 2, right, xLL); // I take the left, she takes the right, or vice versa int xLR; BestMove(left + 1, right - 1, xLR); // I take the right, she takes the right int xRR; BestMove(left, right - 2, xRR); // my worst if I take left int wLeft = *left + std::min(xLL, xLR); // my worst if I take right int wRight = *(r-1) + std::min(xLR, xRR); // the better of the two minGain = std::max(wLeft, wRight); return (wLeft >= wRight)? L: R; } void main() { int pots[] = { 2, 4, 10, 2, 3, 1 }; int minGain; Move m = BestMove(a, a + 6, minGain); }Dynamic programming public class GoldPots { public static int max(int a, int b) { return a < b ? b : a; } public static int solve(int[] a, int player) { int n = a.length; int[][] m = new int[n][n]; int[] sums = new int[n]; sums[0] = a[0]; for(int i = 1; i < n; i++) { sums[i] = sums[i-1] + a[i]; } for(int i = 0; i < n; i++) { m[i][i] = a[i]; } int sumij; for(int k = 1; k < n - 1; k++) { for(int i = 0; i < n - k; i++) { for(int j = i + k; j < n; j++) { sumij = sums[j] - sums[i] + a[i]; m[i][j] = max(sumij - m[i + 1][j], sumij - m[i][j-1]); } } } return m[0][n-1]; } public static void main(String[] args) { int[] a = {1,9,5,3,6,2}; System.out.println(solve(a,0)); } }