Problem: 2050E - Three Strings
Language: java

Solution 1 (runtime: 343 ms):
import java.util.Arrays;
import java.util.Scanner;

public class ThreeStrings {
    public static void main(String[] args) {
        Scanner sc= new Scanner(System.in);
        int t= sc.nextInt();
        for (int i = 0; i < t; i++) {
            String a = sc.next();
            String b = sc.next();
            String c = sc.next();

            int lenA = a.length();
            int lenB = b.length();
            int lenC = c.length();

            long dp[][] = new long[lenA + 10][lenB + 10];

            for(long [] x : dp ) Arrays.fill( x , -1 );

            System.out.println(rec( 0 , 0 , dp , lenA , lenB , a , b, c));
        }
    }

    static long rec ( int x , int y , long dp[][] , int lenA , int lenB , String a , String b, String c){

        if ( x == lenA ){
            long add = 0;
            for(int i = y ; i < lenB ;i++){
                if ( b.charAt(i) != c.charAt( x + i )) add ++ ;
            }
            return add ;
        }
        if ( y == lenB ){
            long add = 0;
            for(int i = x ; i < lenA ;i++){
                if ( a.charAt(i) != c.charAt( i + y )) add ++ ;
            }
            return add ;
        }
        if ( dp[x][y] != -1) return dp[x][y];
        long ans1 = (a.charAt( x ) == c.charAt( x + y ) ? 0 : 1 ) + rec( x + 1 , y , dp , lenA , lenB , a, b, c);
        long ans2 = (b.charAt( y ) == c.charAt( x + y ) ? 0 : 1 ) + rec( x  , y + 1 , dp , lenA , lenB , a, b, c);

        return dp[x][y] = Math.min(ans1 , ans2);
    }
}


Solution 2 (runtime: 280 ms):
import java.util.*;

public class ThreeStrings {
    // Custom print functions
    public static void print(Object print) {
        System.out.print(print);
    }

    public static void println(Object print) {
        System.out.println(print);
    }

    public static void println() {
        System.out.println();
    }

    // Declare a 2D dp array to store memoization results
    static int[][] dp;

    // Function to compute the minimum changes required
    public static int func(int i, int j, char[] a, char[] b, char[] c) {
        int n = c.length;
        // Base case: if all characters are matched
        if (i + j >= n) return 0;

        // Check if result is already calculated
        if (dp[i][j] != -1) return dp[i][j];

        int take_a = Integer.MAX_VALUE;
        int take_b = Integer.MAX_VALUE;

        // Try taking a character from 'a'
        if (i < a.length) {
            if (a[i] == c[i + j]) {
                take_a = func(i + 1, j, a, b, c); // No change needed
            } else {
                take_a = 1 + func(i + 1, j, a, b, c); // Change needed
            }
        }

        // Try taking a character from 'b'
        if (j < b.length) {
            if (b[j] == c[i + j]) {
                take_b = func(i, j + 1, a, b, c); // No change needed
            } else {
                take_b = 1 + func(i, j + 1, a, b, c); // Change needed
            }
        }

        // Store and return the minimum changes required
        return dp[i][j] = Math.min(take_a, take_b);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(); // Number of test cases

        for (int l = 0; l < t; l++) {
            char[] a = sc.next().toCharArray(); // Input string a
            char[] b = sc.next().toCharArray(); // Input string b
            char[] c = sc.next().toCharArray(); // Target string c

            int m = a.length;
            int n = b.length;
            int len = c.length;

            // Initialize dp array with -1
            dp = new int[m + 1][n + 1];
            for (int[] row : dp) {
                Arrays.fill(row, -1);
            }

            // Compute and print result
            println(func(0, 0, a, b, c));
        }
        sc.close();
    }
}