Longest Common Subsequence

Medium Dynamic Programming

Find the length of the longest subsequence common to two strings using 2-D dynamic programming.

Sample input

a = "abcde", b = "ace"

Sample output

Solution

def lcs(a, b):
    m, n = len(a), len(b)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if a[i - 1] == b[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
    return dp[m][n]

print(lcs("abcde", "ace"))

function lcs(a, b) {
  const m = a.length, n = b.length;
  const dp = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (a[i - 1] === b[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
      else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
    }
  }
  return dp[m][n];
}

console.log(lcs("abcde", "ace"));

public class Main {
    static int lcs(String a, String b) {
        int m = a.length(), n = b.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (a.charAt(i - 1) == b.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m][n];
    }

    public static void main(String[] args) {
        System.out.println(lcs("abcde", "ace"));
    }
}

fun lcs(a: String, b: String): Int {
    val m = a.length
    val n = b.length
    val dp = Array(m + 1) { IntArray(n + 1) }
    for (i in 1..m) {
        for (j in 1..n) {
            dp[i][j] = if (a[i - 1] == b[j - 1]) dp[i - 1][j - 1] + 1
            else maxOf(dp[i - 1][j], dp[i][j - 1])
        }
    }
    return dp[m][n]
}

fun main() {
    println(lcs("abcde", "ace"))
}

func lcs(_ a: String, _ b: String) -> Int {
    let ac = Array(a), bc = Array(b)
    let m = ac.count, n = bc.count
    var dp = Array(repeating: Array(repeating: 0, count: n + 1), count: m + 1)
    if m >= 1 && n >= 1 {
        for i in 1...m {
            for j in 1...n {
                if ac[i - 1] == bc[j - 1] {
                    dp[i][j] = dp[i - 1][j - 1] + 1
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
                }
            }
        }
    }
    return dp[m][n]
}

print(lcs("abcde", "ace"))

int lcs(String a, String b) {
  final m = a.length;
  final n = b.length;
  final dp = List.generate(m + 1, (_) => List.filled(n + 1, 0));
  for (var i = 1; i <= m; i++) {
    for (var j = 1; j <= n; j++) {
      if (a[i - 1] == b[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
      } else {
        dp[i][j] = dp[i - 1][j] > dp[i][j - 1] ? dp[i - 1][j] : dp[i][j - 1];
      }
    }
  }
  return dp[m][n];
}

void main() {
  print(lcs("abcde", "ace"));
}

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

int lcs(const string& a, const string& b) {
    int m = a.size(), n = b.size();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (a[i - 1] == b[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
            else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
        }
    }
    return dp[m][n];
}

int main() {
    cout << lcs("abcde", "ace") << endl;
    return 0;
}

#include <stdio.h>
#include <string.h>

int lcs(const char *a, const char *b) {
    int m = strlen(a), n = strlen(b);
    int dp[100][100] = {{0}};
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (a[i - 1] == b[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
            else dp[i][j] = dp[i - 1][j] > dp[i][j - 1] ? dp[i - 1][j] : dp[i][j - 1];
        }
    }
    return dp[m][n];
}

int main() {
    printf("%d\n", lcs("abcde", "ace"));
    return 0;
}