LeetCode 2466: Count Ways To Build Good Strings Solution

Master LeetCode problem 2466 (Count Ways To Build Good Strings), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2466. Count Ways To Build Good Strings

Medium

Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming to build a solution iteratively. We create a 3D DP array where dp[i][j][k] represents the number of ways to build a good string of length i using j zeros and k ones. We iterate through the possible values for i, j, and k to calculate the number of ways to build good strings that meet the constraints.

At each step, we consider the two choices we have: appending a '0' or a '1'. We update the DP array based on these choices and the constraints provided. Finally, we sum up all valid combinations to get the total number of ways to build good strings within the specified length range.

Time complexity: O(N * low * one) where N is the difference between high and low.
Space complexity: O(N * low * one).

Solution Code

class Solution {
    public int countGoodStrings(int low, int high, int zero, int one) {
        int MOD = 1000000007;
        int n = high - low + 1;
        int[][][] dp = new int[n + 1][zero + 1][one + 1];
        
        dp[0][0][0] = 1;
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= zero; j++) {
                for (int k = 0; k <= one; k++) {
                    for (int z = 0; z <= j; z++) {
                        for (int o = 0; o <= k; o++) {
                            if (z <= j && o <= k && z + o > 0) {
                                dp[i][j][k] = (dp[i][j][k] + dp[i - 1][j - z][k - o]) % MOD;
                            }
                        }
                    }
                }
            }
        }
        
        int ans = 0;
        for (int j = 0; j <= zero; j++) {
            for (int k = 0; k <= one; k++) {
                ans = (ans + dp[n][j][k]) % MOD;
            }
        }
        
        return ans;
    }
}

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