# LeetCode 1915. Number of Wonderful Substrings

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:
- "aba" -> "a"
- "aba" -> "b"
- "aba" -> "a"
- "aba" -> "aba"


Example 2:

Input: word = "aabb"
Output: 9
Explanation: The nine wonderful substrings are underlined below:
- "aabb" -> "a"
- "aabb" -> "aa"
- "aabb" -> "aab"
- "aabb" -> "aabb"
- "aabb" -> "a"
- "aabb" -> "abb"
- "aabb" -> "b"
- "aabb" -> "bb"
- "aabb" -> "b"


Example 3:

Input: word = "he"
Output: 2
Explanation: The two wonderful substrings are underlined below:
- "he" -> "h"
- "he" -> "e"

Constraints:

• 1 <= word.length <= 105
• word consists of lowercase English letters from 'a' to 'j'.

To count the number of fulfilling substrings, we need to maintain the prefix count.

Since the word only consist 'a' to 'j' that is 10 letters, then we can use a state of then diigts to note the previous state. each digit dentoes how many previous letter occurs % 2.

Then there could be 1 << 10 different states.

Use count to maintain all the state counts.

count[0] = 1 for no letter, the count is 1. Then later "aa" occurs, its state is "000000000", accumulate the count[0] to result.

Since it is at most one letter, there are two types.

One is all the even freq. res += count[state]. state are the same for this substring

The other is flip one letter. from a to j, try flip its corresponding index.

Time Complexity: O(n). n = word.length().

Space: O(1).

AC Java:

 1 class Solution {
2     public long wonderfulSubstrings(String word) {
3         long[] count = new long[1 << 10];
4         count[0] = 1;
5         int state = 0;
6         long res = 0;
7         int n = word.length();
8         for(int i = 0; i < n; i++){
9             int val = word.charAt(i) - 'a';
10             state ^= (1 << val); // state[i]
11             res += count[state]; // all letter freq are even
12             for(int k = 0; k < 10; k++){
13                 int stateOdd = state ^ (1 << k); // flip one letter
14                 res += count[stateOdd];
15             }
16
17             count[state]++;
18         }
19
20         return res;
21     }
22 }

posted @ 2024-05-14 08:57  Dylan_Java_NYC  阅读(6)  评论(0编辑  收藏  举报