2478. Number of Beautiful Partitions

Description

You are given a string s that consists of the digits '1' to '9' and two integers k and minLength.

A partition of s is called beautiful if:

s is partitioned into k non-intersecting substrings.
Each substring has a length of at least minLength.
Each substring starts with a prime digit and ends with a non-prime digit. Prime digits are '2', '3', '5', and '7', and the rest of the digits are non-prime.

Return the number of beautiful partitions of s. Since the answer may be very large, return it modulo 10⁹ + 7.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "23542185131", k = 3, minLength = 2
Output: 3
Explanation: There exists three ways to create a beautiful partition:
"2354 | 218 | 5131"
"2354 | 21851 | 31"
"2354218 | 51 | 31"

Example 2:

Input: s = "23542185131", k = 3, minLength = 3
Output: 1
Explanation: There exists one way to create a beautiful partition: "2354 | 218 | 5131".

Example 3:

Input: s = "3312958", k = 3, minLength = 1
Output: 1
Explanation: There exists one way to create a beautiful partition: "331 | 29 | 58".

Constraints:

1 <= k, minLength <= s.length <= 1000
s consists of the digits '1' to '9'.

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the number of schemes for dividing the first $i$ characters into $j$ sections. Initialize $f[0][0] = 1$, and the rest $f[i][j] = 0$.

First, we need to determine whether the $i$th character can be the last character of the $j$th section, it needs to meet the following conditions simultaneously:

The $i$th character is a non-prime number;
The $i+1$th character is a prime number, or the $i$th character is the last character of the entire string.

If the $i$th character cannot be the last character of the $j$th section, then $f[i][j]=0$. Otherwise, we have:

$$ f[i][j]=\sum_{t=0}^{i-minLength}f[t][j-1] $$

That is to say, we need to enumerate which character is the end of the previous section. Here we use the prefix sum array $g[i][j] = \sum_{t=0}^{i}f[t][j]$ to optimize the time complexity of enumeration.

Then we have:

$$ f[i][j]=g[i-minLength][j-1] $$

The time complexity is $O(n \times k)$, and the space complexity is $O(n \times k)$. Where $n$ and $k$ are the length of the string $s$ and the number of sections to be divided, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def beautifulPartitions(self, s: str, k: int, minLength: int) -> int:
        primes = '2357'
        if s[0] not in primes or s[-1] in primes:
            return 0
        mod = 10**9 + 7
        n = len(s)
        f = [[0] * (k + 1) for _ in range(n + 1)]
        g = [[0] * (k + 1) for _ in range(n + 1)]
        f[0][0] = g[0][0] = 1
        for i, c in enumerate(s, 1):
            if i >= minLength and c not in primes and (i == n or s[i] in primes):
                for j in range(1, k + 1):
                    f[i][j] = g[i - minLength][j - 1]
            for j in range(k + 1):
                g[i][j] = (g[i - 1][j] + f[i][j]) % mod
        return f[n][k]

class Solution {
    public int beautifulPartitions(String s, int k, int minLength) {
        int n = s.length();
        if (!prime(s.charAt(0)) || prime(s.charAt(n - 1))) {
            return 0;
        }
        int[][] f = new int[n + 1][k + 1];
        int[][] g = new int[n + 1][k + 1];
        f[0][0] = 1;
        g[0][0] = 1;
        final int mod = (int) 1e9 + 7;
        for (int i = 1; i <= n; ++i) {
            if (i >= minLength && !prime(s.charAt(i - 1)) && (i == n || prime(s.charAt(i)))) {
                for (int j = 1; j <= k; ++j) {
                    f[i][j] = g[i - minLength][j - 1];
                }
            }
            for (int j = 0; j <= k; ++j) {
                g[i][j] = (g[i - 1][j] + f[i][j]) % mod;
            }
        }
        return f[n][k];
    }

    private boolean prime(char c) {
        return c == '2' || c == '3' || c == '5' || c == '7';
    }
}

class Solution {
public:
    int beautifulPartitions(string s, int k, int minLength) {
        int n = s.size();
        auto prime = [](char c) {
            return c == '2' || c == '3' || c == '5' || c == '7';
        };
        if (!prime(s[0]) || prime(s[n - 1])) return 0;
        vector<vector<int>> f(n + 1, vector<int>(k + 1));
        vector<vector<int>> g(n + 1, vector<int>(k + 1));
        f[0][0] = g[0][0] = 1;
        const int mod = 1e9 + 7;
        for (int i = 1; i <= n; ++i) {
            if (i >= minLength && !prime(s[i - 1]) && (i == n || prime(s[i]))) {
                for (int j = 1; j <= k; ++j) {
                    f[i][j] = g[i - minLength][j - 1];
                }
            }
            for (int j = 0; j <= k; ++j) {
                g[i][j] = (g[i - 1][j] + f[i][j]) % mod;
            }
        }
        return f[n][k];
    }
};

func beautifulPartitions(s string, k int, minLength int) int {
    prime := func(c byte) bool {
        return c == '2' || c == '3' || c == '5' || c == '7'
    }
    n := len(s)
    if !prime(s[0]) || prime(s[n-1]) {
        return 0
    }
    const mod int = 1e9 + 7
    f := make([][]int, n+1)
    g := make([][]int, n+1)
    for i := range f {
        f[i] = make([]int, k+1)
        g[i] = make([]int, k+1)
    }
    f[0][0], g[0][0] = 1, 1
    for i := 1; i <= n; i++ {
        if i >= minLength && !prime(s[i-1]) && (i == n || prime(s[i])) {
            for j := 1; j <= k; j++ {
                f[i][j] = g[i-minLength][j-1]
            }
        }
        for j := 0; j <= k; j++ {
            g[i][j] = (g[i-1][j] + f[i][j]) % mod
        }
    }
    return f[n][k]
}

function beautifulPartitions(s: string, k: number, minLength: number): number {
    const prime = (c: string): boolean => {
        return c === '2' || c === '3' || c === '5' || c === '7';
    };

    const n: number = s.length;
    if (!prime(s[0]) || prime(s[n - 1])) return 0;

    const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(k + 1).fill(0));
    const g: number[][] = new Array(n + 1).fill(0).map(() => new Array(k + 1).fill(0));
    const mod: number = 1e9 + 7;

    f[0][0] = g[0][0] = 1;

    for (let i = 1; i <= n; ++i) {
        if (i >= minLength && !prime(s[i - 1]) && (i === n || prime(s[i]))) {
            for (let j = 1; j <= k; ++j) {
                f[i][j] = g[i - minLength][j - 1];
            }
        }
        for (let j = 0; j <= k; ++j) {
            g[i][j] = (g[i - 1][j] + f[i][j]) % mod;
        }
    }

    return f[n][k];
}

2478. Number of Beautiful Partitions

Description

Solutions

Solution 1: Dynamic Programming

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