Data Structures & Algorithms Interview Questions

Chen J. - "Leetcode 347: Heap + Hashtable Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer

Kavithadevi P. - "hash maps work in key value pair. The keys are hashed with a hash algorithm and resulting hashcode(integer) with related value are stored. Accessing a value, removing an element, Searching the hash map: 1) The hash map value can be accessed in O(1) time once you know the key. 2) If the key is not known, the hashmap value can be accessed in O(n) since you have to iterate atleast once. "See full answer

Saurabh S. - "Basic Approach As BST inorder traversal will result in a sequence of increasing order. Store that order in a vector and get the k-1 index to get the Kth smallest element, similarly access the N-K+1 th element will be the Kth largest element Time Complexity: O(n) Space Complexity O(n) Space Optimized Approach For Kth smallest , start inorder traversal, and keep a counter, decrement the counter when you access the node element. When the counter turns 0 that elementwill be the Kth smal"See full answer

Rasika D. - "A doubly linked list is like a train where the engine is its head, and the compartments doors are interconnected, you can travel though them. The last compartment is the tail and is not connected with anything. Following is a visual: Head C1 C2 C3....Cn Tail Pseudocode: **class Node: Data pointer next pointer prev End class DLL = NULL //pointer to list insertion: if DLL is NULL: head = new Node() DLL = head DLL.prev = NU"See full answer

Mark S. - "int a_array[10] = {3,6,4,7,2,1,9}; int index = 0; int index2 = 0; for ( index = 0; index < sizeof(a_array); index++ ) { int tmpindex = index + 1; if ( tmpindex <= sizeof(a_array) ) { for ( index2 = tmpindex; index2 < sizeof(a_array); index2++ ) { if ( aarray[index] <= aarray[index2] ) { print( "%d is the NGE of %d" array[index2], array[index]); break; "See full answer

Rishabh N. - "I first asked few clarifying questions like the return array may need not contain the list of building in the same order, to which the interviewer agreed. Then I came up with an approach where we iterate the array from right to left and keep a max variable which will keep the value of the current max. When we find an item which is greater than max we update the max and add this element into our solution. The interviewer agreed for the approach. I discussed few corner scenarios with the interview"See full answer

Iris F. - "#include #include #include using namespace std; void printComs(int prev, int start, int end, int target) { if (start >= end) return; while (start target) { end--; } else { st"See full answer

Mourya C. - "I try to solve this initially using quick select where will take a pivot element and position the remaining elements and check if the current index is answer or not and continue the same but it requires o(n*n), but interviewee is expecting the best from me, so at the end i tried solving using heaps where will check the difference between k and n-k to use min or max heap after that we will heap the array, and will keep popping the element k-1 and return the peek one which leads to answer."See full answer

Shivam P. - " First, sort the array in ascending order. This ensures that we can easily check the triangle inequality condition. Use a loop to iterate through the array. For each triplet of consecutive elements, check if they satisfy the triangle inequality condition a+b>ca+b>c. As soon as you find a valid tuple, return it. If no valid tuple is found, return null. This approach is efficient with a time complexity of O(nlog⁡n)O(nlogn) due to the sorting step, followed by a linear scan of the array"See full answer