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2514. Count Anagrams

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

• Java
• C++
• Python
• Go

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

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