Software Engineer Data Structures & Algorithms Interview Questions

Weida H. - "filter function usually exists in some high level programming that adopt FP paradigm. It taks a sequence of items and a predicate function, and returns (conceptually) a subset of the items that satisfy the predicate. Adopt this kind of operation (filter, map, reduce, take, pairwise ...) can help writting cleaner code, and reduce the usage of mutable part in the program, lower the possibility of making human mistake. Take Python for example (although const-ness is not exists in Python), assu"See full answer

Yitayal B. - "import re def naturalsortkey(s): Split the string into a list of text and numbers "item10" -> ["item", 10] return [int(text) if text.isdigit() else text.lower() for text in re.split(r'(\d+)', s)] def natural_sort(arr): return sorted(arr, key=naturalsortkey) \# Example Usage data = ["item10", "item2", "item1", "item20", "2nd item", "1st item"] sorteddata = naturalsort(data) print(f"Standard Sort: {sorted(data)}") print(f"Natural Sort: {sorted_data}") "See full answer

Rudransh V. - "I got it right"See full answer

Alessandro R. - "import java.util.*; public class NetworkTopology { public int topologytype(int N, int M, int[] input3, int[] input4) { if (M != N - 1 && M != N) return -1; // Fast check for invalid cases int[] degree = new int[N + 1]; // Degree of each node (1-based indexing) // Build the degree array for (int i = 0; i < M; i++) { degree[input3[i]]++; degree[input4[i]]++; } // Check for Bus Topology boolean isBus = (M"See full answer

Sourav K. - "We can use a pool of memory where each index we can store the parent of current index then we can use same lca approach for array instead of pointers."See full answer

Anonymous Eel - "String commonStr(String str1, String str2) { int len1 = str1.length(); int len2 = str2.length(); if (len1 == 0 || len2 == 0) return ""; // let dpx reprsent the longest common str of 0...x int dp = new int len1 + 1; int maxLen = 0; int endIndex = 0; for (int i = 1; i <= len1; i++) { for (int j = 1; j <= len2; j++) { if (str1.charAt(i-1) == str2.charAt(j-1)) { dpi = dpi-1 + 1; "See full answer