LeetCode 3129: Find All Possible Stable Binary Arrays I Solution

Master LeetCode problem 3129 (Find All Possible Stable Binary Arrays I), 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.

3129. Find All Possible Stable Binary Arrays I

Medium

Dynamic Programming Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will create a 3D dynamic programming array dp, where dp[i][j][k] represents the number of stable binary arrays with i zeros, j ones, and k as the last element of the array. We will then iterate through all possible values for i, j, and k and calculate the number of stable binary arrays.

Algorithmic Idea

Initialize a 3D DP array dp with dimensions dp[zero+1][one+1][2].
Set the base case dp[0][1][1] = 1, as the only stable binary array with zero zeros and one one is [0,1].
Iterate through all possible values for i, j, and k from 0 to zero, one, and 1 respectively.
Update the dp array based on the following conditions:
- If k is 0, dp[i][j][0] = dp[i-1][j][1] + dp[i][j][0].
- If k is 1, dp[i][j][1] = dp[i][j-1][0] + dp[i][j][1].
Return the sum of all dp[zero][one][0] and dp[zero][one][1] modulo 10^9 + 7.

Time Complexity

The time complexity of this solution is O(zero * one), where zero and one are the given integers.

Space Complexity

The space complexity of this solution is O(zero * one) for the DP array.

Solution Code

class Solution {
    public int findStableBinaryArrays(int zero, int one, int limit) {
        int MOD = 1000000007;
        int[][][] dp = new int[zero+1][one+1][2];
        dp[0][1][1] = 1;
        
        for (int i = 0; i <= zero; i++) {
            for (int j = 0; j <= one; j++) {
                for (int k = 0; k <= 1; k++) {
                    if (k == 0 && i > 0) {
                        dp[i][j][0] = (dp[i-1][j][1] + dp[i][j][0]) % MOD;
                    }
                    if (k == 1 && j > 0) {
                        dp[i][j][1] = (dp[i][j-1][0] + dp[i][j][1]) % MOD;
                    }
                }
            }
        }
        
        return (dp[zero][one][0] + dp[zero][one][1]) % MOD;
    }
}

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