Problem: 2050E - Three Strings
Language: java

Solution 1 (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();
    }
}


Solution 2 (runtime: 327 ms):
import java.util.Scanner;



public class Main{

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        int t = scanner.nextInt();

        scanner.nextLine(); // Consume the newline



        for (int i = 0; i < t; i++) {

            String a = scanner.nextLine();

            String b = scanner.nextLine();

            String c = scanner.nextLine();



            System.out.println(minChanges(a, b, c));

        }

    }



    private static int minChanges(String a, String b, String c) {

        int lenA = a.length();

        int lenB = b.length();

        int lenC = c.length();



        int[][] dp = new int[lenA + 1][lenB + 1];

        for (int i = 0; i <= lenA; i++) {

            for (int j = 0; j <= lenB; j++) {

                dp[i][j] = Integer.MAX_VALUE;

            }

        }

        dp[0][0] = 0;



        for (int i = 0; i <= lenA; i++) {

            for (int j = 0; j <= lenB; 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));

                }

            }

        }



        return dp[lenA][lenB];

    }

}





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