Software Engineer Coding Interview Questions

Anonymous Wombat - "Was the statement very similar to the leetcode or was it changed and only the main idea remained?"See full answer

Guilherme M. - "Question: An array of n integers is given, and a positive integer k, where k << n. k indicates that the absolute difference between each element's current index (icurrent) and the index in the sorted array (isorted) is less than k (|icurr - isorted| < k). Sort the given array. The most common solution is with a Heap: def solution(arr, k): min_heap = [] result = [] for i in range(len(arr)) heapq.heappush(min_heap, arr[i]) "See full answer

Jerry O. - "\# Definition for a binary tree node. class TreeNode: def init(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def maxPathSum(self, root: TreeNode) -> int: self.max_sum = float('-inf')"See full answer

Vidhyadhar V. - "public class CircularBuffer { private T[] buffer; private int head; private int tail; private int size; private final int capacity; public CircularBuffer(int capacity) { this.capacity = capacity; this.buffer = (T[]) new Object[capacity]; this.head = 0; this.tail = 0; this.size = 0; } public void enqueue(T item) { if (isFull()) { throw new IllegalStateException("Buffer is full"); } buf"See full answer

Sarvesh G. - "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"See full answer

Yash N. - "import time class Task: def init\(self, description, interval=None): self.description = description self.interval = interval self.next_run = time.time() class SimpleTaskScheduler: def init\(self): self.tasks = [] def add_task(self, description, interval=None): self.tasks.append(Task(description, interval)) def run(self, duration=60): end_time = time.time() + duration while time.time() < end_time: curr"See full answer

Anonymous Unicorn - "naive solution: def countprefixpairs(words): n = len(words) count = 0 for i in range(n): for j in range(i + 1, n): if words[i].startswith(words[j]) or words[j].startswith(words[i]): count += 1 return count using tries for when the list of words is very long: from collections import Counter class TrieNode: def init(self): self.children = {} self.count = 0 # To count the number of words ending at this node"See full answer

K.nithish K. - "MOD = 10**9 + 7 def max_stability(reliability, availability): max_stability = 1 for r, a in zip(reliability, availability): Compute stability of the current server stability = r * a if stability != 0: Multiply into max_stability and take modulo maxstability = (maxstability * stability) % MOD return max_stability reliability = [1, 2, 2] availability = [1, 1, 3] print(max_stability(reliability, availability)) # Output the result mo"See full answer

Nathan C. - "def mostefficientseqscore(parentheses, efficiencyratings): mes = [] for i in range(len(parentheses)): mes.append((parentheses[i], max(efficiency_ratings[i])) return sum([m[1] for m in mes]) `"See full answer

Nitin P. - "public static void sortBinaryArray(int[] array) { int len = array.length; int[] res = new int[len]; int r=len-1; for (int value : array) { if(value==1){ res[r]= 1; r--; } } System.out.println(Arrays.toString(res)); } `"See full answer

Ugo C. - "function preorderToInorder(preorder) { let inorder = []; let stack = []; let root = preorder[0]; stack.push(root); for (let i = 1; i 0 && stack[stack.length - 1] 0) { root = stack.pop(); inorder.push(r"See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"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

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