Data Structures & Algorithms Interview Questions

Shashwat K. - "Binary Search on the array and after than compare the numbers at low and the high pointers whichever is closest is the answer. Because after the binary search low will be pointing to a number which is immediate greater than x and high will be pointing to a number which is immediate lesser than x. int low = 0; int high = n-1; while(low <= high){ int mid = (low + high) / 2; if(x == arr[mid]) return mid; //if x is already present then it will be the closest else if(x < arr[mid]) high"See full answer

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Tiago R. - "function areSentencesSimilar(sentence1, sentence2, similarPairs) { if (sentence1.length !== sentence2.length) return false; for (let i=0; i (w1 === word1 && !visited.has(w2)) || (w2 === word1 && !visited.has(w1))); if (!edge) { "See full answer

Gaurav K. - "I think sliding window will work here and it is the most optimized approach to solve this question."See full answer

Anonymous Wasp - "I was able to provide the optimal approach and coded it up"See full answer

Vaibhav C. - "To determine if a graph is not a tree, you can check for the following conditions: Presence of cycles: A graph is not a tree if it contains cycles. In a tree, there should be exactly one unique path between any two vertices. If you can find a cycle in the graph, it cannot be a tree. Insufficient number of edges: A tree with N vertices will have exactly N-1 edges. If the graph has fewer or more than N-1 edges, then it is not a tree. Disconnected components: A tree is a connected graph, m"See full answer

Gaddipati V. - "NA"See full answer

Tiago R. - "function knapsack(weights, values, cap) { const indicesByValue = Object.keys(weights).map(weight => parseInt(weight)); indicesByValue.sort((a, b) => values[b]-values[a]); const steps = new Map(); function knapsackStep(cap, sack) { if (steps.has(sack)) { return steps.get(sack); } let maxOutput = 0; for (let index of indicesByValue) { if (!sack.has(index) && weights[index] <= cap) { maxOutput ="See full answer

Rohit B. - "class Solution: def missingNumber(self, nums: list[int]) -> int: Sorting approach n = len(nums) s = n*(n+1)//2 r = s - sum(nums) return self.r l = [3,0,1] print(missingNumber(l))"See full answer

Ugo C. - "function merge(L1, L2) { let L3 = { data: null, next: null }; let prev = L3; while (L1 != null || L2 != null) { if (L1 == null) { prev.next = L2; L2 = L2.next; } else if (L2 == null) { prev.next = L1; L1 = L1.next; } else if (L1.data < L2.data) { prev.next = L1; L1 = L1.next; } else { prev.next = L2; L2 = L2.next; } prev = prev.next; } return L3.next; }"See full answer

Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

Tiago R. - "function visitChildren(node) { let leftSubtreeHeight = 0; let rightSubtreeHeight = 0; let isChildrenBalanced = true; if (node.left) { const { isBalanced, height } = visitChildren(node.left); isChildrenBalanced = isChildrenBalanced && isBalanced; leftSubtreeHeight += height + 1; } if (isChildrenBalanced && node.right) { const { isBalanced, height } = visitChildren(node.right); isChildrenBalanced = isChildrenBalanced && isBalan"See full answer

Tiago R. - "function spiralCopy(inputMatrix) { if (inputMatrix.length === 1) return inputMatrix[0]; let lowerY = 0; let upperY = inputMatrix.length-1; let lowerX = 0; let upperX = inputMatrix[0].length-1; const output = []; while (true) { if (lowerX > upperX) break; for (let x = lowerX; x upperY) break; for (let y = lowerY; y <= upperY; y++) { output.push(inputMatrix[y][u"See full answer

Ugo C. - "function countRotations(arr, low, high) { if (high low && arr[mid] arr[mid]) { return countRotations(arr, low, mid - 1); } return countRotations(arr, mid + 1, high); } "See full answer

Alex M. - "A red-black tree is a self-balancing binary search tree. The motivation for this is that the benefits of O(logN) search, insertion, and deletion that a binary tree provides us will disappear if we let the tree get too "imbalanced" (e.g. there are too many nodes on one side of the tree or some branches have a depth that is way out of proportion to the average branch depth). This imbalance will occur if we don't adjust the tree after inserting or deleting nodes, hence our need for self-balancing c"See full answer

Rohan M. - "My solution is simple; it does an in-order DFS traversal to create an array of in-order elements then it searches through the array to find the node we want the successor of. finally we return the node that is 1 after the input node, in the case our input node is the last element of our DFS we know there is no successor, therefore it returns None/null. CODE INSTRUCTIONS: 1) The method fi"See full answer