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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 $j \leq \textit{limit}$, otherwise return $0$.
  • If $j = 0$, return $1$ when $k = 0$ and $i \leq \textit{limit}$, 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 $\textit{limit}$ $0$s in the subarray, i.e., the situation where the $\textit{limit} + 1$
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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

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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];
    }
}

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class Solution {
public:
    int numberOfStableArrays(int zero, int one, int limit) {
        const int mod = 1e9 + 7;
        using ll = long long;
        vector<vector<array<ll, 2>>> f = vector<vector<array<ll, 2>>>(zero + 1, vector<array<ll, 2>>(one + 1, {-1, -1}));
        auto dfs = [&](auto&& dfs, int i, int j, int k) -> ll {
            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(dfs, i - 1, j, 0) + dfs(dfs, i - 1, j, 1) - dfs(dfs, i - limit - 1, j, 1) + mod) % mod;
            } else {
                res = (dfs(dfs, i, j - 1, 0) + dfs(dfs, i, j - 1, 1) - dfs(dfs, i, j - limit - 1, 0) + mod) % mod;
            }
            return res;
        };
        return (dfs(dfs, zero, one, 0) + dfs(dfs, zero, one, 1)) % mod;
    }
};

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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
}

Solution 2: Dynamic Programming

We can also convert the memoization search of Solution 1 into dynamic programming.

We define $f[i][j][k]$ as the number of stable binary arrays using $i$ $0$s and $j$ $1$s, and the last number is $k$. So the answer is $f[zero][one][0] + f[zero][one][1]$.

Initially, we have $f[i][0][0] = 1$, where $1 \leq i \leq \min(\textit{limit}, \textit{zero})$; and $f[0][j][1] = 1$, where $1 \leq j \leq \min(\textit{limit}, \textit{one})$.

The state transition equation is as follows:

  • $f[i][j][0] = f[i - 1][j][0] + f[i - 1][j][1] - f[i - \textit{limit} - 1][j][1]$.
  • $f[i][j][1] = f[i][j - 1][0] + f[i][j - 1][1] - f[i][j - \textit{limit} - 1][0]$.

The time complexity is $O(zero \times one)$, and the space complexity is $O(zero \times one)$.

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class Solution:
    def numberOfStableArrays(self, zero: int, one: int, limit: int) -> int:
        mod = 10**9 + 7
        f = [[[0, 0] for _ in range(one + 1)] for _ in range(zero + 1)]
        for i in range(1, min(limit, zero) + 1):
            f[i][0][0] = 1
        for j in range(1, min(limit, one) + 1):
            f[0][j][1] = 1
        for i in range(1, zero + 1):
            for j in range(1, one + 1):
                x = 0 if i - limit - 1 < 0 else f[i - limit - 1][j][1]
                y = 0 if j - limit - 1 < 0 else f[i][j - limit - 1][0]
                f[i][j][0] = (f[i - 1][j][0] + f[i - 1][j][1] - x) % mod
                f[i][j][1] = (f[i][j - 1][0] + f[i][j - 1][1] - y) % mod
        return sum(f[zero][one]) % mod

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class Solution {
    public int numberOfStableArrays(int zero, int one, int limit) {
        final int mod = (int) 1e9 + 7;
        long[][][] f = new long[zero + 1][one + 1][2];
        for (int i = 1; i <= Math.min(zero, limit); ++i) {
            f[i][0][0] = 1;
        }
        for (int j = 1; j <= Math.min(one, limit); ++j) {
            f[0][j][1] = 1;
        }
        for (int i = 1; i <= zero; ++i) {
            for (int j = 1; j <= one; ++j) {
                long x = i - limit - 1 < 0 ? 0 : f[i - limit - 1][j][1];
                long y = j - limit - 1 < 0 ? 0 : f[i][j - limit - 1][0];
                f[i][j][0] = (f[i - 1][j][0] + f[i - 1][j][1] - x + mod) % mod;
                f[i][j][1] = (f[i][j - 1][0] + f[i][j - 1][1] - y + mod) % mod;
            }
        }
        return (int) ((f[zero][one][0] + f[zero][one][1]) % mod);
    }
}

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class Solution {
public:
    int numberOfStableArrays(int zero, int one, int limit) {
        const int mod = 1e9 + 7;
        using ll = long long;
        ll f[zero + 1][one + 1][2];
        memset(f, 0, sizeof(f));
        for (int i = 1; i <= min(zero, limit); ++i) {
            f[i][0][0] = 1;
        }
        for (int j = 1; j <= min(one, limit); ++j) {
            f[0][j][1] = 1;
        }
        for (int i = 1; i <= zero; ++i) {
            for (int j = 1; j <= one; ++j) {
                ll x = i - limit - 1 < 0 ? 0 : f[i - limit - 1][j][1];
                ll y = j - limit - 1 < 0 ? 0 : f[i][j - limit - 1][0];
                f[i][j][0] = (f[i - 1][j][0] + f[i - 1][j][1] - x + mod) % mod;
                f[i][j][1] = (f[i][j - 1][0] + f[i][j - 1][1] - y + mod) % mod;
            }
        }
        return (f[zero][one][0] + f[zero][one][1]) % mod;
    }
};

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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 i := 1; i <= min(zero, limit); i++ {
        f[i][0][0] = 1
    }
    for j := 1; j <= min(one, limit); j++ {
        f[0][j][1] = 1
    }
    for i := 1; i <= zero; i++ {
        for j := 1; j <= one; j++ {
            f[i][j][0] = (f[i-1][j][0] + f[i-1][j][1]) % mod
            if i-limit-1 >= 0 {
                f[i][j][0] = (f[i][j][0] - f[i-limit-1][j][1] + mod) % mod
            }
            f[i][j][1] = (f[i][j-1][0] + f[i][j-1][1]) % mod
            if j-limit-1 >= 0 {
                f[i][j][1] = (f[i][j][1] - f[i][j-limit-1][0] + mod) % mod
            }
        }
    }
    return (f[zero][one][0] + f[zero][one][1]) % mod
}

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function numberOfStableArrays(zero: number, one: number, limit: number): number {
    const mod = 1e9 + 7;
    const f: number[][][] = Array.from({ length: zero + 1 }, () =>
        Array.from({ length: one + 1 }, () => [0, 0]),
    );

    for (let i = 1; i <= Math.min(limit, zero); i++) {
        f[i][0][0] = 1;
    }
    for (let j = 1; j <= Math.min(limit, one); j++) {
        f[0][j][1] = 1;
    }

    for (let i = 1; i <= zero; i++) {
        for (let j = 1; j <= one; j++) {
            const x = i - limit - 1 < 0 ? 0 : f[i - limit - 1][j][1];
            const y = j - limit - 1 < 0 ? 0 : f[i][j - limit - 1][0];
            f[i][j][0] = (f[i - 1][j][0] + f[i - 1][j][1] - x + mod) % mod;
            f[i][j][1] = (f[i][j - 1][0] + f[i][j - 1][1] - y + mod) % mod;
        }
    }

    return (f[zero][one][0] + f[zero][one][1]) % mod;
}

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