Problem: 2050E - Three Strings
Language: java

Solution 1 (runtime: 374 ms):
import java.util.*;

public class Solution {
    static int dp[][];
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while(t-->0){
            String a=sc.next();
            String b=sc.next();
            String c=sc.next();
            dp=new int[a.length()+1][b.length()+1];

            for(int i=0;i<dp.length;i++){
                Arrays.fill(dp[i],-1);
            }
            System.out.println(solve(a,b,c,0,0,0));

        }
    }
    public static int solve(String a,String b,String c,int i,int j,int k){
        if(i==a.length() || j==b.length()){
            return calc(a,b,c,i,j,k);
        }
        if(dp[i][j]!=-1) return dp[i][j];
        if(c.charAt(k)==a.charAt(i) && c.charAt(k)==b.charAt(j)){
            return dp[i][j]=Math.min(solve(a,b,c,i+1,j,k+1),solve(a,b,c,i,j+1,k+1));

        }
        else if(c.charAt(k)==a.charAt(i) ){
            return dp[i][j]=Math.min(solve(a,b,c,i+1,j,k+1),1+solve(a,b,c,i,j+1,k+1));
        }
        else if(c.charAt(k)==b.charAt(j)){
            return dp[i][j]=Math.min(solve(a,b,c,i,j+1,k+1),1+solve(a,b,c,i+1,j,k+1));
        }
        else return dp[i][j]=1+Math.min(solve(a,b,c,i+1,j,k+1),solve(a,b,c,i,j+1,k+1));
    }

    public static int calc(String a,String b,String c,int i,int j,int k){
        int count=0;
        if(i==a.length()){
            while(j<b.length()){
                if(b.charAt(j)!=c.charAt(k)){
                    count++;
                }
                k++;
                j++;
            }

        }
        else{
            while(i<a.length()){
                if(a.charAt(i)!=c.charAt(k)){
                    count++;
                }
                i++;
                k++;
            }
        }
        return count;
    }
}

Solution 2 (runtime: 233 ms):
import java.util.Scanner;

public class Main {
    private static final int MAX_LEN = 2003; // Ensure larger array size for safety

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        scanner.nextLine(); // Consume the newline character

        while (t-- > 0) {
            String a = scanner.nextLine();
            String b = scanner.nextLine();
            String c = scanner.nextLine();

            int n = a.length();
            int m = b.length();
            int[][] dp = new int[n + 1][m + 1]; // Use n + 1 and m + 1 for proper indexing

            for (int i = 0; i <= n; i++) {
                for (int j = 0; j <= m; j++) {
                    dp[i][j] = MAX_LEN; // Initialize with a large value
                }
            }

            dp[0][0] = 0;

            for (int i = 0; i <= n; i++) {
                for (int j = 0; j <= m; j++) {
                    if (i > 0) {
                        dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + (a.charAt(i - 1) != c.charAt(i + j - 1) ? 1 : 0));
                    }
                    if (j > 0) {
                        dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + (b.charAt(j - 1) != c.charAt(i + j - 1) ? 1 : 0));
                    }
                }
            }

            System.out.println(dp[n][m]);
        }
        scanner.close(); // Close the scanner to avoid resource leaks
    }
}