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: 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();
    }
}