Amazon Coding Interview Questions

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

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

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

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

Firdous A. - "Approach (BFS + Horizontal Distance) Assign a horizontal distance (HD) to each node. Root → HD = 0 Left child → HD = parent HD - 1 Right child → HD = parent HD + 1 Do a BFS (level order traversal). If a node with a given HD is seen for the first time, add it to the result. Ignore later nodes with the same HD (because only the top one is visible). After traversal, sort by HD and print nodes left to righ"See full answer

Anonymous Goat - "Batch Packing Problem In Amazon’s massive warehouse inventory, there are different types of products. You are given an array products of size n, where products[i] represents the number of items of product type i. These products need to be packed into batches for shipping. The batch packing must adhere to the following conditions: No two items in the same batch can be of the same product type. The number of items packed in the current batch must be strictly greater than the number pack"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer

Khushbu R. - "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"See full answer

Chris R. - "It was like say we have a library A which has a library B as a dependency and so on, how would we determine in the dependency chain that whether there is a circular depedency?"See full answer

Dinesh K. - "i dont know"See full answer

Jeff S. - " class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function diameterOfTree(root) { if (root === null || root.left === null & root.right === null) { return 0; } function countBranch(node, count) { if (node.left === null && node.right === null) { return count; } let left = node.left === null ? 0 : countBranch(node.left, count+1); let right = no"See full answer

Teddy Y. - "class Node: def init(self, value): self.value = value self.children = [] def inorder_traversal(root): if not root: return [] result = [] n = len(root.children) for i in range(n): result.extend(inorder_traversal(root.children[i])) if i == n // 2: result.append(root.value) if n == 0: result.append(root.value) return result Example usage: root = Node(1) child1 = Node(2) chil"See full answer

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"See full answer

Gundala tarun,cse2020 V. - "#include // Naive method to find a pair in an array with a given sum void findPair(int nums[], int n, int target) { // consider each element except the last for (int i = 0; i < n - 1; i++) { // start from the i'th element until the last element for (int j = i + 1; j < n; j++) { // if the desired sum is found, print it if (nums[i] + nums[j] == target) { printf("Pair found (%d, %d)\n", nums[i], nums[j]); return; } } } // we reach here if the pair is not found printf("Pair not found"); } "See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

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

Anonymous Unicorn - "input_logs = [ f"{senderid} {receiverid} {transaction_count}" "1 2 2", "3 2 42", "2 2 22", "1 1 12", "2 1 1", "2 5 4", "4 2 15" ] input_threshold = 20 exptected_output = [ list of user_ids that made more than 20 transactions sorted by number of transactions in descending order "3", # 42 transactions "2", # 27 transactions (22 + 1 + 4) #"4", # 15 transactions #"1" # 14 transactions (2 + 12 + 1) ] def gettopapi_users(logs, thres"See full answer