Data Structures & Algorithms Interview Questions

Ramachandra N. - "def sortedSquares(nums): n = len(nums) result = [0] * n left, right = 0, n - 1 pos = n - 1 while left abs(nums[right]): result[pos] = nums[left] ** 2 left += 1 else: result[pos] = nums[right] ** 2 right -= 1 pos -= 1 return result `"See full answer

Ramachandra N. - "from collections import deque from typing import List def longestsubarraydifflessthan_n(nums: List[int], N: int) -> int: """ Find the length of the longest contiguous subarray such that the difference between any two elements in the subarray is less than N. Equivalent condition: max(subarray) - min(subarray) < N Approach (Optimal): Sliding window with two monotonic deques: max_d: decreasing deque of indices (front is index of current max"See full answer

Ramachandra N. - " def closest_palindrome(n: str) -> str: """ Finds the closest palindromic number to n (excluding itself). Assumptions: If two palindromes are equally close, return the smaller one. n is a positive integer represented as a string. Time Complexity: O(1) Space Complexity: O(1) """ length = len(n) num = int(n) Helper to build palindrome from a prefix def makepalindrome(prefix: int, isodd_length: bool) -> int: s = str(prefi"See full answer

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Venkata rakesh M. - "Here if we breakdown each dependency [A,B] , We need to check if there a cycle in Dependency Graph. If there is cycle installation is not possible, If there is no cycle installation is possible. Steps : 1: Build the graph 2: Perform DFS based Cycle Detection 3: Check each package if those packages have cycle or not."See full answer

Maykon henrique D. - "function longestCommonPrefix(arr1, arr2) { const prefixSet = new Set(); for (let num of arr1) { let str = num.toString(); for (let i = 1; i <= str.length; i++) { prefixSet.add(str.substring(0, i)); } } let longestPrefix = ""; for (let num of arr2) { let str = num.toString(); for (let i = 1; i <= str.length; i++) { let prefix = str.substring(0, i); if (prefixSet.has(prefix)) { "See full answer

Yeshwanth D. - "Was this for an entry level engineer role?"See full answer

Sh R. - "i responded using a multi sourced BFS and in place marking, then i checked the final grid to see if any free spots were left unmarked."See full answer

Richard W. - "int getMaxWater(vector& nums) { int n = nums.size(); int mx = INT_MIN; int i=0, j=n-1; while(i<j) { int water = (j - i) * min(nums[i], nums[j]); mx = max(mx, water); if(nums[i] < nums[j]){ i++; } else { j--; } } return mx; } `"See full answer

Anonymous Roadrunner - "def is_palindrome(s: str) -> bool: new = '' for a in s: if a.isalpha() or a.isdigit(): new += a.lower() return (new == new[::-1]) debug your code below print(is_palindrome('abcba')) `"See full answer

Rick E. - " O(n) - characters in the string O(n) - stack def identify_adjacent(s: str, k: int) -> str: stack = [] n = len(s) for ch in s: if stack: topch, topcnt = stack[-1] if top_ch == ch: top_cnt += 1 stack[-1] = (ch, top_cnt) else: top_cnt = 1 stack.append((ch,1)) if top_cnt == k: stack.pop() else: stack.append"See full answer

Frederick K. - "\# An program that prints out the peak elements in a list of integers. Pseudocode: 1. Define a function that takes a list of integers as input. 2. Initialize an empty list to store the peak elements. 3. Loop through the list of integers. 4. For each element, check if it is greater than its neighbors. 5. If it is, add it to the list of peak elements. 6. Return the list of peak elements. def findpeakelements(nums): if not nums: return [] peaks = [] n = len(nums"See full answer

Gaurav B. - "class TreeNode(var val: Int, var left: TreeNode? = null, var right: TreeNode? = null) fun isAverageOfDescendants(root: TreeNode?): Boolean { fun helper(node: TreeNode?): Triple { if (node == null) return Triple(0, 0, true) val (leftSum, leftCount, leftValid) = helper(node.left) val (rightSum, rightCount, rightValid) = helper(node.right) val totalSum = leftSum + rightSum val totalCount = leftCount + rightCount // If leaf n"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Balasubramanian R. - "Inorder traversal of the tree should be the solution for this problem."See full answer

Rahul J. - "Constraints: 4-direction moves; no mode switching (pick exactly one of {1=bicycle, 2=bike, 3=car, 4=bus} for the full trip). Per-mode search: If a mode’s per-step time/cost are uniform, run BFS on allowed cells. Then totaltime = steps × timeperstep, tie-break by steps × costper_step. If time/cost vary by cell (given matrices), run Dijkstra per mode minimizing (totaltime, totalcost) lexicographically. Maintain the best ⟨time, cost⟩ per cell; relax when the new pair is strictly better. S"See full answer

Anni P. - "I firstly discuss the brute force approach in O(n^2) time complexity , than i moved to O(nlogn) tine complexity than i discussed the O(n) time complexity and O(n) space complexity . But interviewer want more optimised solution , in O(n) time complexity without using extra space , The solution wants O(1) space complexity i have to do changes in same array without using any space . This method is something like i have to place positive values to its original position by swapping and rest negativ"See full answer

Srikant V. - "Used Recursive approach to traverse the binary search tree and sum the values of the nodes that fall within the specified range [low, high]"See full answer