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Hard
2091
Biweekly Contest 112 Q4
Greedy
Hash Table
Math
String
Combinatorics

2842. Count K-Subsequences of a String With Maximum Beauty

中文文档

Description

You are given a string s and an integer k.

A k-subsequence is a subsequence of s, having length k, and all its characters are unique, i.e., every character occurs once.

Let f(c) denote the number of times the character c occurs in s.

The beauty of a k-subsequence is the sum of f(c) for every character c in the k-subsequence.

For example, consider s = "abbbdd" and k = 2:

  • f('a') = 1, f('b') = 3, f('d') = 2
  • Some k-subsequences of s are:
    • "abbbdd" -> "ab" having a beauty of f('a') + f('b') = 4
    • "abbbdd" -> "ad" having a beauty of f('a') + f('d') = 3
    • "abbbdd" -> "bd" having a beauty of f('b') + f('d') = 5

Return an integer denoting the number of k-subsequences whose beauty is the maximum among all k-subsequences. Since the answer may be too large, return it modulo 109 + 7.

A subsequence of a string is a new string formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.

Notes

  • f(c) is the number of times a character c occurs in s, not a k-subsequence.
  • Two k-subsequences are considered different if one is formed by an index that is not present in the other. So, two k-subsequences may form the same string.

 

Example 1:

Input: s = "bcca", k = 2
Output: 4
Explanation: From s we have f('a') = 1, f('b') = 1, and f('c') = 2.
The k-subsequences of s are: 
bcca having a beauty of f('b') + f('c') = 3 
bcca having a beauty of f('b') + f('c') = 3 
bcca having a beauty of f('b') + f('a') = 2 
bcca having a beauty of f('c') + f('a') = 3
bcca having a beauty of f('c') + f('a') = 3 
There are 4 k-subsequences that have the maximum beauty, 3. 
Hence, the answer is 4.

Example 2:

Input: s = "abbcd", k = 4
Output: 2
Explanation: From s we have f('a') = 1, f('b') = 2, f('c') = 1, and f('d') = 1. 
The k-subsequences of s are: 
abbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5
abbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5 
There are 2 k-subsequences that have the maximum beauty, 5. 
Hence, the answer is 2.

 

Constraints:

  • 1 <= s.length <= 2 * 105
  • 1 <= k <= s.length
  • s consists only of lowercase English letters.

Solutions

Solution 1: Greedy + Combinatorial Mathematics

First, we use a hash table $f$ to count the occurrence of each character in the string $s$, i.e., $f[c]$ represents the number of times character $c$ appears in the string $s$.

Since a $k$-subsequence is a subsequence of length $k$ in the string $s$ with unique characters, if the number of different characters in $f$ is less than $k$, then there is no $k$-subsequence, and we can directly return $0$.

Otherwise, to maximize the beauty value of the $k$-subsequence, we need to make characters with high beauty values appear as much as possible in the $k$-subsequence. Therefore, we can sort the values in $f$ in reverse order to get an array $vs$.

We denote the occurrence of the $k$th character in the array $vs$ as $val$, and there are $x$ characters with an occurrence of $val$.

Then we first find out the characters with occurrences greater than $val$, multiply the occurrences of each character to get the initial answer $ans$, and update the remaining number of characters to be selected to $k$. We need to select $k$ characters from $x$ characters, so the answer needs to be multiplied by the combination number $C_x^k$, and finally multiplied by $val^k$, i.e., $ans = ans \times C_x^k \times val^k$.

Note that we need to use fast power and modulo operations here.

The time complexity is $O(n)$, and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string, and $\Sigma$ is the character set. In this problem, the character set is lowercase letters, so $|\Sigma| = 26$.

Python3

class Solution:
    def countKSubsequencesWithMaxBeauty(self, s: str, k: int) -> int:
        f = Counter(s)
        if len(f) < k:
            return 0
        mod = 10**9 + 7
        vs = sorted(f.values(), reverse=True)
        val = vs[k - 1]
        x = vs.count(val)
        ans = 1
        for v in vs:
            if v == val:
                break
            k -= 1
            ans = ans * v % mod
        ans = ans * comb(x, k) * pow(val, k, mod) % mod
        return ans

Java

class Solution {
    private final int mod = (int) 1e9 + 7;

    public int countKSubsequencesWithMaxBeauty(String s, int k) {
        int[] f = new int[26];
        int n = s.length();
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (++f[s.charAt(i) - 'a'] == 1) {
                ++cnt;
            }
        }
        if (cnt < k) {
            return 0;
        }
        Integer[] vs = new Integer[cnt];
        for (int i = 0, j = 0; i < 26; ++i) {
            if (f[i] > 0) {
                vs[j++] = f[i];
            }
        }
        Arrays.sort(vs, (a, b) -> b - a);
        long ans = 1;
        int val = vs[k - 1];
        int x = 0;
        for (int v : vs) {
            if (v == val) {
                ++x;
            }
        }
        for (int v : vs) {
            if (v == val) {
                break;
            }
            --k;
            ans = ans * v % mod;
        }
        int[][] c = new int[x + 1][x + 1];
        for (int i = 0; i <= x; ++i) {
            c[i][0] = 1;
            for (int j = 1; j <= i; ++j) {
                c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
            }
        }
        ans = ((ans * c[x][k]) % mod) * qpow(val, k) % mod;
        return (int) ans;
    }

    private long qpow(long a, int n) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countKSubsequencesWithMaxBeauty(string s, int k) {
        int f[26]{};
        int cnt = 0;
        for (char& c : s) {
            if (++f[c - 'a'] == 1) {
                ++cnt;
            }
        }
        if (cnt < k) {
            return 0;
        }
        vector<int> vs(cnt);
        for (int i = 0, j = 0; i < 26; ++i) {
            if (f[i]) {
                vs[j++] = f[i];
            }
        }
        sort(vs.rbegin(), vs.rend());
        const int mod = 1e9 + 7;
        long long ans = 1;
        int val = vs[k - 1];
        int x = 0;
        for (int v : vs) {
            x += v == val;
        }
        for (int v : vs) {
            if (v == val) {
                break;
            }
            --k;
            ans = ans * v % mod;
        }
        int c[x + 1][x + 1];
        memset(c, 0, sizeof(c));
        for (int i = 0; i <= x; ++i) {
            c[i][0] = 1;
            for (int j = 1; j <= i; ++j) {
                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
            }
        }
        auto qpow = [&](long long a, int n) {
            long long ans = 1;
            for (; n; n >>= 1) {
                if (n & 1) {
                    ans = ans * a % mod;
                }
                a = a * a % mod;
            }
            return ans;
        };
        ans = (ans * c[x][k] % mod) * qpow(val, k) % mod;
        return ans;
    }
};

Go

func countKSubsequencesWithMaxBeauty(s string, k int) int {
	f := [26]int{}
	cnt := 0
	for _, c := range s {
		f[c-'a']++
		if f[c-'a'] == 1 {
			cnt++
		}
	}
	if cnt < k {
		return 0
	}
	vs := []int{}
	for _, x := range f {
		if x > 0 {
			vs = append(vs, x)
		}
	}
	sort.Slice(vs, func(i, j int) bool {
		return vs[i] > vs[j]
	})
	const mod int = 1e9 + 7
	ans := 1
	val := vs[k-1]
	x := 0
	for _, v := range vs {
		if v == val {
			x++
		}
	}
	for _, v := range vs {
		if v == val {
			break
		}
		k--
		ans = ans * v % mod
	}
	c := make([][]int, x+1)
	for i := range c {
		c[i] = make([]int, x+1)
		c[i][0] = 1
		for j := 1; j <= i; j++ {
			c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod
		}
	}
	qpow := func(a, n int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	ans = (ans * c[x][k] % mod) * qpow(val, k) % mod
	return ans
}

TypeScript

function countKSubsequencesWithMaxBeauty(s: string, k: number): number {
    const f: number[] = new Array(26).fill(0);
    let cnt = 0;
    for (const c of s) {
        const i = c.charCodeAt(0) - 97;
        if (++f[i] === 1) {
            ++cnt;
        }
    }
    if (cnt < k) {
        return 0;
    }
    const mod = BigInt(10 ** 9 + 7);
    const vs: number[] = f.filter(v => v > 0).sort((a, b) => b - a);
    const val = vs[k - 1];
    const x = vs.filter(v => v === val).length;
    let ans = 1n;
    for (const v of vs) {
        if (v === val) {
            break;
        }
        --k;
        ans = (ans * BigInt(v)) % mod;
    }
    const c: number[][] = new Array(x + 1).fill(0).map(() => new Array(k + 1).fill(0));
    for (let i = 0; i <= x; ++i) {
        c[i][0] = 1;
        for (let j = 1; j <= Math.min(i, k); ++j) {
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % Number(mod);
        }
    }
    const qpow = (a: bigint, n: number): bigint => {
        let ans = 1n;
        for (; n; n >>>= 1) {
            if (n & 1) {
                ans = (ans * a) % BigInt(mod);
            }
            a = (a * a) % BigInt(mod);
        }
        return ans;
    };
    ans = (((ans * BigInt(c[x][k])) % mod) * qpow(BigInt(val), k)) % mod;
    return Number(ans);
}