There are n red balls and m blue balls Find no of arrangements of balls in which not more than k balls of same color are placed next to each other in Java | Hackerrank | LeetCode

public class BallArrangements {
    public static int countArrangements(int n, int m, int k) {
        int[][][] dp = new int[n + 1][m + 1][k + 1];

        // Initialize base cases
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                dp[i][j][0] = 1;
            }
        }

        // Fill the dp array
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                for (int l = 1; l <= k; l++) {
                    // If we place a red ball at position i
                    dp[i][j][0] += dp[i - 1][j][l];

                    // If we place a blue ball at position j
                    dp[i][j][0] += dp[i][j - 1][l];

                    // If we place a red ball at position i and there are less than k reds
                    dp[i][j][l] += dp[i - 1][j][l - 1];

                    // If we place a blue ball at position j and there are less than k blues
                    dp[i][j][l] += dp[i][j - 1][l - 1];
                }
            }
        }

        int totalArrangements = 0;
        for (int l = 0; l <= k; l++) {
            totalArrangements += dp[n][m][l];
        }

        return totalArrangements;
    }

    public static void main(String[] args) {
        int n = 3; // Number of red balls
        int m = 2; // Number of blue balls
        int k = 1; // Maximum consecutive balls of the same color allowed
        int arrangements = countArrangements(n, m, k);
        System.out.println("Total arrangements: " + arrangements);
    }
}