Problem: 2050E - Three Strings
Language: java

Solution 1 (runtime: 390 ms):
import java.util.*;

public class Q1 {
    public static long solve(long n, HashMap<Long, Long> map) {
        if (n <= 3) {
            return 1;
        }

        if (map.containsKey(n)) {
            return map.get(n);
        }

        long count = solve(n / 4, map) + solve(n / 4, map);
        map.put(n, count);
        return map.get(n);
    }

    public static long gcd(long a, long b) {
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // Base case
        if (a == b)
            return a;

        // a is greater
        if (a > b) {
            if (a % b == 0)
                return b;
            return gcd(a - b, b);
        }
        if (b % a == 0)
            return a;
        return gcd(a, b - a);
    }

    public static int solve(int ai, int ci, String a, String b, String c, int dp[][]) {
        int bi = ci - ai;
        if (ai == a.length()) {
            String s1 = b.substring(bi);
            String s2 = c.substring(ci);
            if (s1.equals(s2)) {
                return 0;
            }
            int cnt = 0;
            for (int i = 0; i < s1.length(); i++) {
                if (s1.charAt(i) != s2.charAt(i)) {
                    cnt++;
                }
            }

            return cnt;
        }

        if (bi == b.length()) {
            String s1 = a.substring(ai);
            String s2 = c.substring(ci);
            if (s1.equals(s2)) {
                return 0;
            }
            int cnt = 0;
            for (int i = 0; i < s1.length(); i++) {
                if (s1.charAt(i) != s2.charAt(i)) {
                    cnt++;
                }
            }

            return cnt;
        }

        if (dp[ai][ci] != -1) {
            return dp[ai][ci];
        }

        int min = Integer.MAX_VALUE;
        if (a.charAt(ai) == c.charAt(ci)) {
            min = Math.min(solve(ai + 1, ci + 1, a, b, c, dp), min);
        }

        if (b.charAt(bi) == c.charAt(ci)) {
            min = Math.min(solve(ai, ci + 1, a, b, c, dp), min);
        }

        if (a.charAt(ai) != c.charAt(ci) || b.charAt(bi) != c.charAt(ci)) {
            min = Math.min(1 + Math.min(solve(ai + 1, ci + 1, a, b, c, dp), solve(ai, ci + 1, a, b, c, dp)), min);
        }

        return dp[ai][ci] = min;
    }

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

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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