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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2501_2600.s2514_count_anagrams;

// #Hard #String #Hash_Table #Math #Counting #Combinatorics
// #2023_03_21_Time_22_ms_(100.00%)_Space_43_MB_(98.15%)

import java.util.Arrays;

/**
 * 2514 - Count Anagrams\.
 *
 * Hard
 *
 * 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 i^th word of `t` is a **permutation** of the i^th 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.
**/
public class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countAnagrams(String s) {
        var charArray = s.toCharArray();
        long ans = 1L;
        long mul = 1L;
        var cnt = new int[26];
        int j = 0;
        for (char c : charArray) {
            if (c == ' ') {
                Arrays.fill(cnt, 0);
                j = 0;
            } else {
                mul = mul * ++cnt[c - 'a'] % MOD;
                ans = ans * ++j % MOD;
            }
        }
        return (int) (ans * pow(mul, MOD - 2) % MOD);
    }

    private long pow(long x, int n) {
        var res = 1L;
        for (; n > 0; n /= 2) {
            if (n % 2 > 0) {
                res = res * x % MOD;
            }
            x = x * x % MOD;
        }
        return res;
    }
}