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3130. Find All Possible Stable Binary Arrays II

Description

You are given 3 positive integers zero, one, and limit.

A binary array arr is called stable if:

  • The number of occurrences of 0 in arr is exactly zero.
  • The number of occurrences of 1 in arr is exactly one.
  • Each subarray of arr with a size greater than limit must contain both 0 and 1.

Return the total number of stable binary arrays.

Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

Input: zero = 1, one = 1, limit = 2

Output: 2

Explanation:

The two possible stable binary arrays are [1,0] and [0,1].

Example 2:

Input: zero = 1, one = 2, limit = 1

Output: 1

Explanation:

The only possible stable binary array is [1,0,1].

Example 3:

Input: zero = 3, one = 3, limit = 2

Output: 14

Explanation:

All the possible stable binary arrays are [0,0,1,0,1,1], [0,0,1,1,0,1], [0,1,0,0,1,1], [0,1,0,1,0,1], [0,1,0,1,1,0], [0,1,1,0,0,1], [0,1,1,0,1,0], [1,0,0,1,0,1], [1,0,0,1,1,0], [1,0,1,0,0,1], [1,0,1,0,1,0], [1,0,1,1,0,0], [1,1,0,0,1,0], and [1,1,0,1,0,0].

 

Constraints:

  • 1 <= zero, one, limit <= 1000

Solutions

We design a function $dfs(i, j, k)$ to represent the number of stable binary arrays that satisfy the problem conditions when there are $i$ $0$s and $j$ $1$s left, and the next number to be filled is $k$. The answer is $dfs(zero, one, 0) + dfs(zero, one, 1)$.

The calculation process of the function $dfs(i, j, k)$ is as follows:

  • If $i < 0$ or $j < 0$, return $0$.
  • If $i = 0$, return $1$ when $k = 1$ and $j \leq \text{limit}$, otherwise return $0$.
  • If $j = 0$, return $1$ when $k = 0$ and $i \leq \text{limit}$, otherwise return $0$.
  • If $k = 0$, we consider the case where the previous number is $0$, $dfs(i - 1, j, 0)$, and the case where the previous number is $1$, $dfs(i - 1, j, 1)$. If the previous number is $0$, it may cause more than $\text{limit}$ $0$s in the subarray, i.e., the situation where the $\text{limit} + 1$
  • class Solution {
    private final int mod = (int) 1e9 + 7;
    private Long[][][] f;
    private int limit;

    public int numberOfStableArrays(int zero, int one, int limit) {
        f = new Long[zero + 1][one + 1][2];
        this.limit = limit;
        return (int) ((dfs(zero, one, 0) + dfs(zero, one, 1)) % mod);
    }

    private long dfs(int i, int j, int k) {
        if (i < 0 || j < 0) {
            return 0;
        }
        if (i == 0) {
            return k == 1 && j <= limit ? 1 : 0;
        }
        if (j == 0) {
            return k == 0 && i <= limit ? 1 : 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        if (k == 0) {
            f[i][j][k]
                = (dfs(i - 1, j, 0) + dfs(i - 1, j, 1) - dfs(i - limit - 1, j, 1) + mod) % mod;
        } else {
            f[i][j][k]
                = (dfs(i, j - 1, 0) + dfs(i, j - 1, 1) - dfs(i, j - limit - 1, 0) + mod) % mod;
        }
        return f[i][j][k];
    }
}

  • using ll = long long;

class Solution {
public:
    int numberOfStableArrays(int zero, int one, int limit) {
        this->limit = limit;
        f = vector<vector<array<ll, 2>>>(zero + 1, vector<array<ll, 2>>(one + 1, {-1, -1}));
        return (dfs(zero, one, 0) + dfs(zero, one, 1)) % mod;
    }

private:
    const int mod = 1e9 + 7;
    int limit;
    vector<vector<array<ll, 2>>> f;

    ll dfs(int i, int j, int k) {
        if (i < 0 || j < 0) {
            return 0;
        }
        if (i == 0) {
            return k == 1 && j <= limit;
        }
        if (j == 0) {
            return k == 0 && i <= limit;
        }
        ll& res = f[i][j][k];
        if (res != -1) {
            return res;
        }
        if (k == 0) {
            res = (dfs(i - 1, j, 0) + dfs(i - 1, j, 1) - dfs(i - limit - 1, j, 1) + mod) % mod;
        } else {
            res = (dfs(i, j - 1, 0) + dfs(i, j - 1, 1) - dfs(i, j - limit - 1, 0) + mod) % mod;
        }
        return res;
    }
};

  • class Solution:
    def numberOfStableArrays(self, zero: int, one: int, limit: int) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if i == 0:
                return int(k == 1 and j <= limit)
            if j == 0:
                return int(k == 0 and i <= limit)
            if k == 0:
                return (
                    dfs(i - 1, j, 0)
                    + dfs(i - 1, j, 1)
                    - (0 if i - limit - 1 < 0 else dfs(i - limit - 1, j, 1))
                )
            return (
                dfs(i, j - 1, 0)
                + dfs(i, j - 1, 1)
                - (0 if j - limit - 1 < 0 else dfs(i, j - limit - 1, 0))
            )

        mod = 10**9 + 7
        ans = (dfs(zero, one, 0) + dfs(zero, one, 1)) % mod
        dfs.cache_clear()
        return ans


  • func numberOfStableArrays(zero int, one int, limit int) int {
	const mod int = 1e9 + 7
	f := make([][][2]int, zero+1)
	for i := range f {
		f[i] = make([][2]int, one+1)
		for j := range f[i] {
			f[i][j] = [2]int{-1, -1}
		}
	}
	var dfs func(i, j, k int) int
	dfs = func(i, j, k int) int {
		if i < 0 || j < 0 {
			return 0
		}
		if i == 0 {
			if k == 1 && j <= limit {
				return 1
			}
			return 0
		}
		if j == 0 {
			if k == 0 && i <= limit {
				return 1
			}
			return 0
		}
		res := &f[i][j][k]
		if *res != -1 {
			return *res
		}
		if k == 0 {
			*res = (dfs(i-1, j, 0) + dfs(i-1, j, 1) - dfs(i-limit-1, j, 1) + mod) % mod
		} else {
			*res = (dfs(i, j-1, 0) + dfs(i, j-1, 1) - dfs(i, j-limit-1, 0) + mod) % mod
		}
		return *res
	}
	return (dfs(zero, one, 0) + dfs(zero, one, 1)) % mod
}

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