Skip to content
Programmingoneonone
Programmingoneonone

Learn everything about programming

  • Home
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
    • Cybersecurity
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone
Programmingoneonone

Learn everything about programming

HackerRank The Longest Common Subsequence problem solution

YASH PAL, 31 July 202430 November 2025

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.

HackerRank The Longest Common Subsequence problem solution

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 AlgorithmsHackerRank

Post navigation

Previous post
Next post

Are you a student and stuck with your career or worried about real-time things, and don't know how to manage your learning phase? Which profession to choose? and how to learn new things according to your goal, and land a dream job. Then this might help to you.

Hi My name is YASH PAL, founder of this Blog and a Senior Software engineer with 5+ years of Industry experience. I personally helped 40+ students to make a clear goal in their professional lives. Just book a one-on-one personal call with me for 30 minutes for 300 Rupees. Ask all your doubts and questions related to your career to set a clear roadmap for your professional life.

Book session - https://wa.me/qr/JQ2LAS7AASE2M1

Pages

  • About US
  • Contact US
  • Privacy Policy

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes