README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2514.Count%20Anagrams/README_EN.md	2069	Biweekly Contest 94 Q4	Hash Table Math String Combinatorics Counting

2514. Count Anagrams

中文文档

Description

You are given a string s containing one or more words. Every consecutive pair of words is separated by a single space ' '.

A string t is an anagram of string s if the ith word of t is a permutation of the ith word of s.

• For example, "acb dfe" is an anagram of "abc def", but "def cab" and "adc bef" are not.

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

 

Example 1:

Input: s = "too hot"
Output: 18
Explanation: Some of the anagrams of the given string are "too hot", "oot hot", "oto toh", "too toh", and "too oht".

Example 2:

Input: s = "aa"
Output: 1
Explanation: There is only one anagram possible for the given string.

 

Constraints:

• 1 <= s.length <= 105
• s consists of lowercase English letters and spaces ' '.
• There is single space between consecutive words.

Solutions

Solution 1

Python3

mod = 10**9 + 7
f = [1]
for i in range(1, 10**5 + 1):
    f.append(f[-1] * i % mod)


class Solution:
    def countAnagrams(self, s: str) -> int:
        ans = 1
        for w in s.split():
            cnt = Counter(w)
            ans *= f[len(w)]
            ans %= mod
            for v in cnt.values():
                ans *= pow(f[v], -1, mod)
                ans %= mod
        return ans

Java

import java.math.BigInteger;

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

    public int countAnagrams(String s) {
        int n = s.length();
        long[] f = new long[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            f[i] = f[i - 1] * i % MOD;
        }
        long p = 1;
        for (String w : s.split(" ")) {
            int[] cnt = new int[26];
            for (int i = 0; i < w.length(); ++i) {
                ++cnt[w.charAt(i) - 'a'];
            }
            p = p * f[w.length()] % MOD;
            for (int v : cnt) {
                p = p * BigInteger.valueOf(f[v]).modInverse(BigInteger.valueOf(MOD)).intValue()
                    % MOD;
            }
        }
        return (int) p;
    }
}

C++

class Solution {
public:
    const int mod = 1e9 + 7;

    int countAnagrams(string s) {
        stringstream ss(s);
        string w;
        long ans = 1, mul = 1;
        while (ss >> w) {
            int cnt[26] = {0};
            for (int i = 1; i <= w.size(); ++i) {
                int c = w[i - 1] - 'a';
                ++cnt[c];
                ans = ans * i % mod;
                mul = mul * cnt[c] % mod;
            }
        }
        return ans * pow(mul, mod - 2) % mod;
    }

    long pow(long x, int n) {
        long res = 1L;
        for (; n; n /= 2) {
            if (n % 2) res = res * x % mod;
            x = x * x % mod;
        }
        return res;
    }
};

Go

const mod int = 1e9 + 7

func countAnagrams(s string) int {
	ans, mul := 1, 1
	for _, w := range strings.Split(s, " ") {
		cnt := [26]int{}
		for i, c := range w {
			i++
			cnt[c-'a']++
			ans = ans * i % mod
			mul = mul * cnt[c-'a'] % mod
		}
	}
	return ans * pow(mul, mod-2) % mod
}

func pow(x, n int) int {
	res := 1
	for ; n > 0; n >>= 1 {
		if n&1 > 0 {
			res = res * x % mod
		}
		x = x * x % mod
	}
	return res
}

Solution 2

Python3

class Solution:
    def countAnagrams(self, s: str) -> int:
        mod = 10**9 + 7
        ans = mul = 1
        for w in s.split():
            cnt = Counter()
            for i, c in enumerate(w, 1):
                cnt[c] += 1
                mul = mul * cnt[c] % mod
                ans = ans * i % mod
        return ans * pow(mul, -1, mod) % mod