LeetCode 3130: Find All Possible Stable Binary Arrays II Solution
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3130. Find All Possible Stable Binary Arrays II
Problem Explanation
Explanation
To solve this problem, we can use dynamic programming. We will keep track of the number of stable binary arrays that can be formed up to a certain point in the array. We will iterate through all possible states of the array, considering the number of zeros, ones, and the last subarray length.
Here's the algorithm:
- Initialize a 3D array
dpto store the number of stable arrays for each state. - Iterate through all possible states, updating
dpbased on the rules mentioned in the problem. - Return the total number of stable binary arrays modulo 10^9 + 7.
Time complexity: O(zero * one * limit)
Space complexity: O(zero * one * limit)
Solution Code
class Solution {
public int numOfArrays(int zero, int one, int limit) {
int MOD = 1000000007;
int[][][] dp = new int[zero + 1][one + 1][limit + 1];
for (int i = 1; i <= one; i++) {
dp[0][i][1] = 1;
}
for (int i = 1; i <= zero; i++) {
for (int j = 0; j <= one; j++) {
for (int k = 1; k <= limit; k++) {
long sum = 0;
for (int l = 1; l <= j; l++) {
sum = (sum + dp[i - 1][j - l][k - 1]) % MOD;
}
dp[i][j][k] = (int) sum;
}
}
}
int result = 0;
for (int i = 0; i <= one; i++) {
result = (result + dp[zero][i][limit]) % MOD;
}
return result;
}
}Try It Yourself
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