HackerRank The Longest Common Subsequence problem solution YASH PAL, 31 July 2024 In this HackerRank The Longest Common Subsequence problem solution You have given two sequences of integers, A = [a[1], … , a[n]] and B = [b[1],….,b[m]], find the longest common subsequence and print it as a line of space-separated integers. If there are multiple common subsequences with the same maximum length, print any one of them. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++.Problem solution in C. Problem solution in Python. #!/bin/python3 import sys # parse input n,m = input().strip().split(' ') n = int(n) m = int(m) A = [int(x) for x in input().strip().split(' ')] B = [int(x) for x in input().strip().split(' ')] # initialize table for DP table = dict() for i in range(0,n+1): loc = tuple([i,0]) table[loc] = list() for j in range(0,m+1): loc = tuple([0,j]) table[loc] = list() # populate table for i in range(1,n+1): for j in range(1,m+1): a = A[i-1] b = B[j-1] if a == b: seq = table[tuple([i-1,j-1])].copy() seq.append(a) table[tuple([i,j])] = seq else: ti = table[tuple([i-1,j])].copy() tj = table[tuple([i,j-1])].copy() if len(ti) > len(tj): table[tuple([i,j])] = ti.copy() else: table[tuple([i,j])] = tj.copy() # output result out = table[tuple([n,m])] out = ' '.join(str(x) for x in out) print(out) {“mode”:”full”,”isActive”:false} Problem solution in Java. import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); int m = scanner.nextInt(); int[] a = new int[n]; for (int i = 0; i < n; ++i) { a[i] = scanner.nextInt(); } int[] b = new int[m]; for (int i = 0; i < m; ++i) { b[i] = scanner.nextInt(); } int[][] opt = new int[n + 1][m + 1]; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= m; ++j) { if (a[i - 1] == b[j - 1]) { opt[i][j] = opt[i - 1][j - 1] + 1; } else { opt[i][j] = Math.max(opt[i][j - 1], opt[i - 1][j]); } } } //System.out.println(opt[n][m]); int i = n, j = m; Stack<Integer> stack = new Stack<>(); while (i > 0 && j > 0) { if (a[i - 1] == b[j - 1]) { stack.push(a[i - 1]); i -= 1; j -= 1; } else if (opt[i][j - 1] >= opt[i - 1][j]) { j -= 1; } else if (opt[i][j - 1] < opt[i - 1][j]) { i -= 1; } } StringBuilder sb = new StringBuilder(); while (!stack.isEmpty()) { sb.append(stack.pop() + " "); } System.out.println(sb.toString().trim()); } } {“mode”:”full”,”isActive”:false} Problem solution in C++. #include <ios> #include <iostream> int dp[105][105] = {}; int parenti[105][105] = {}; int parentj[105][105] = {}; int arr1[105] = {}; int arr2[105] = {}; int n, m; bool first = true; int lcs(int i, int j) { if (i == -1) return 0; else if (j == -1) return 0; else if (dp[i][j] == -1) { if (arr1[i] == arr2[j]) { dp[i][j] = lcs(i-1, j-1)+1; parenti[i][j] = i-1; parentj[i][j] = j-1; } else { if (lcs(i-1, j) > lcs(i, j-1)) { dp[i][j] = lcs(i-1, j); parenti[i][j] = i-1; parentj[i][j] = j; } else { dp[i][j] = lcs(i, j-1); parenti[i][j] = i; parentj[i][j] = j-1; } } } return dp[i][j]; } void print(int i, int j) { if (i != -1 && j != -1) { print(parenti[i][j], parentj[i][j]); if (arr1[i] == arr2[j]) { if (first) first = false; else std::cout << " "; std::cout << arr1[i]; } } } int main() { std::ios_base::sync_with_stdio(false); for (int i = 0; i < 105; i++) { for (int j = 0; j < 105; j++) { dp[i][j] = -1; parenti[i][j] = -1; parentj[i][j] = -1; } } std::cin >> n >> m; for (int i = 0; i < n; i++) std::cin >> arr1[i]; for (int i = 0; i < m; i++) std::cin >> arr2[i]; lcs(n-1, m-1); print(n-1, m-1); } {“mode”:”full”,”isActive”:false} Problem solution in C. #include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int* longestCommonSubsequence(int a_size, int* a, int b_size, int* b, int *result_size) { int memo[a_size + 1][b_size + 1]; // initialize table. for (int i = 0; i <= a_size; i++) memo[i][0] = 0; for (int j = 1; j <= b_size; j++) memo[0][j] = 0; // Find the length of the longest substring for (int i = 1, r = 0; i <= a_size; i++, r++) for (int j = 1, c = 0; j <= b_size; j++, c++) if (a[r] == b[c]) memo[i][j] = memo[i - 1][j - 1] + 1; else memo[i][j] = (memo[i - 1][j] > memo[i][j - 1]) ? memo[i - 1][j] : memo[i][j - 1]; // Read the memo backwards to find the LCS int len = memo[a_size][b_size]; int r = a_size; int c = b_size; int* result = (int*)malloc(sizeof(int) * len); while (len) { if (a[r-1] == b[c-1]) { --len; --r; --c; result[len] = a[r]; } else if (memo[r - 1][c] == len) --r; else --c; } *result_size = memo[a_size][b_size]; return result; } int main() { int n; int m; scanf("%i %i", &n, &m); int *a = malloc(sizeof(int) * n); for (int a_i = 0; a_i < n; a_i++) { scanf("%i",&a[a_i]); } int *b = malloc(sizeof(int) * m); for (int b_i = 0; b_i < m; b_i++) { scanf("%i",&b[b_i]); } int result_size; int* result = longestCommonSubsequence(n, a, m, b, &result_size); for(int result_i = 0; result_i < result_size; result_i++) { if(result_i) { printf(" "); } printf("%d", result[result_i]); } puts(""); return 0; } {“mode”:”full”,”isActive”:false} Algorithms coding problems solutions