[LeetCode 1371] Find the Longest Substring Containing Vowels in Even Counts

Given the string s, return the size of the longest substring containing each vowel an even number of times. That is, 'a', 'e', 'i', 'o', and 'u' must appear an even number of times.

Example 1:

Input: s = "eleetminicoworoep"
Output: 13
Explanation: The longest substring is "leetminicowor" which contains two each of the vowels: e, i and o and zero of the vowels: a and u.

Example 2:

Input: s = "leetcodeisgreat"
Output: 5
Explanation: The longest substring is "leetc" which contains two e's.

Example 3:

Input: s = "bcbcbc"
Output: 6
Explanation: In this case, the given string "bcbcbc" is the longest because all vowels: a, e, i, o and u appear zero times.

Constraints:

• 1 <= s.length <= 5 x 10^5
• s contains only lowercase English letters.

Key observations:

1.  For a given substring, we only care if all of its vowel counts are even. 

2. Each time we process a new vowel character C, its count parity flips. 

3. Even - Even = Even; Odd - Odd = Odd; Even - Odd = Odd; Odd - Even = Odd.

Using the above observations, we derive the following algorithm.

1.  Use bitmask to track s[0, i]'s vowel parity information.

2. Each time we process a vowel, update the running bitmask by flipping this vowel's parity.

3. For each different parity state, we save its earliest occurence in a map. For each state, if it has never appeared before, save it to the map. If it has appeared before, update the global best result. A repeating state means the substring in between its earliest and current occurence has all even counts of vowels. (Observation No. 3)

class Solution {
    public int findTheLongestSubstring(String s) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int best = 0, curr = 0;
        for(int i = 0; i < s.length(); i++) {
            if(s.charAt(i) == 'a' || s.charAt(i) == 'e' || s.charAt(i) == 'i'
              || s.charAt(i) == 'o' || s.charAt(i) == 'u') {
                int diff = s.charAt(i) - 'a';
                curr ^= (1 << diff);
            }
            map.putIfAbsent(curr, i);
            best = Math.max(best, i - map.get(curr));
        }
        return best;
    }
}

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