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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1901_2000.s1915_number_of_wonderful_substrings;

// #Medium #String #Hash_Table #Bit_Manipulation #Prefix_Sum
// #2022_05_14_Time_31_ms_(82.46%)_Space_54.6_MB_(38.60%)

/**
 * 1915 - Number of Wonderful Substrings\.
 *
 * Medium
 *
 * A **wonderful** string is a string where **at most one** letter appears an **odd** number of times.
 *
 * *   For example, `"ccjjc"` and `"abab"` are wonderful, but `"ab"` is not.
 *
 * Given a string `word` that consists of the first ten lowercase English letters (`'a'` through `'j'`), return _the **number of wonderful non-empty substrings** in_ `word`_. If the same substring appears multiple times in_ `word`_, then count **each occurrence** separately._
 *
 * A **substring** is a contiguous sequence of characters in a string.
 *
 * **Example 1:**
 *
 * **Input:** word = "aba"
 *
 * **Output:** 4
 *
 * **Explanation:** The four wonderful substrings are underlined below: 
 *
 * - "**a**ba" -> "a" 
 *
 * - "a**b**a" -> "b" 
 *
 * - "ab**a**" -> "a" 
 *
 * - "**aba**" -> "aba"
 *
 * **Example 2:**
 *
 * **Input:** word = "aabb"
 *
 * **Output:** 9
 *
 * **Explanation:** The nine wonderful substrings are underlined below: 
 *
 * - "**a**abb" -> "a" 
 *
 * - "**aa**bb" -> "aa" 
 *
 * - "**aab**b" -> "aab" 
 *
 * - "**aabb**" -> "aabb" 
 *
 * - "a**a**bb" -> "a" 
 *
 * - "a**abb**" -> "abb" 
 *
 * - "aa**b**b" -> "b" 
 *
 * - "aa**bb**" -> "bb" 
 *
 * - "aab**b**" -> "b"
 *
 * **Example 3:**
 *
 * **Input:** word = "he"
 *
 * **Output:** 2
 *
 * **Explanation:** The two wonderful substrings are underlined below: 
 *
 * - "**h**e" -> "h" 
 *
 * - "h**e**" -> "e"
 *
 * **Constraints:**
 *
 * *   1 <= word.length <= 10⁵
 * *   `word` consists of lowercase English letters from `'a'` to `'j'`.
**/
public class Solution {
    public long wonderfulSubstrings(String word) {
        int[] count = new int[1024];
        long res = 0;
        int cur = 0;
        count[0] = 1;
        for (int i = 0; i < word.length(); i++) {
            cur ^= 1 << (word.charAt(i) - 'a');
            res += count[cur];
            for (int j = 0; j < 10; j++) {
                res += count[cur ^ (1 << j)];
            }
            ++count[cur];
        }
        return res;
    }
}