LeetCode 2272. Substring With Largest Variance

原题链接在这里：https://leetcode.com/problems/substring-with-largest-variance/

题目：

The variance of a string is defined as the largest difference between the number of occurrences of any 2 characters present in the string. Note the two characters may or may not be the same.

Given a string s consisting of lowercase English letters only, return the largest variance possible among all substrings of s.

A substring is a contiguous sequence of characters within a string. 

Example 1:

Input: s = "aababbb"
Output: 3
Explanation:
All possible variances along with their respective substrings are listed below:
- Variance 0 for substrings "a", "aa", "ab", "abab", "aababb", "ba", "b", "bb", and "bbb".
- Variance 1 for substrings "aab", "aba", "abb", "aabab", "ababb", "aababbb", and "bab".
- Variance 2 for substrings "aaba", "ababbb", "abbb", and "babb".
- Variance 3 for substring "babbb".
Since the largest possible variance is 3, we return it.

Example 2:

Input: s = "abcde"
Output: 0
Explanation:
No letter occurs more than once in s, so the variance of every substring is 0.

Constraints:

• 1 <= s.length <= 104
• s consists of lowercase English letters.

题解：

The variance is the difference between largest char occurance - smallest char occurance.

We could mark most apparence char as 1, lest appearence char as -1.

aaaab would be like [1, 1, 1, 1, -1]. Then this question would be like maximum subarray.

However it needs to have at least one b since aaaa is not a valid case.

How to make sure there is at least one b, update the result when count of b is larger than 0.

When count of a is already smaller than count of b, we need to reset count of a and count of b as 0. However, it must make sure there is more b in the rest string.

Since baaa, at the first char b, count of b is 1, count of a is 0. Should we reset? No. If we reset there would never be a b in the rest and aaa is not a solid case.

Time Complexity: O(n). 26 * 26 * n. n = s.length().

Space: O(1). 

AC Java:

class Solution {
    public int largestVariance(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        int [] map = new int[26];
        for(int i = 0; i < s.length(); i++){
            map[s.charAt(i) - 'a']++;
        }
        
        int res = 0;
        for(int i = 0; i < 26; i++){
            for(int j = 0; j < 26; j++){
                if(map[i] == 0 || map[j] == 0 || i == j){
                    continue;
                }
                
                int countI = 0;
                int countJ = 0;
                int remainJ = map[j];
                for(int k = 0; k < s.length(); k++){
                    int ind = (int)(s.charAt(k) - 'a');
                    if(ind == i){
                        countI++;
                    }else if(ind == j){
                        countJ++;
                        remainJ--;
                    }
                    
                    if(countI > countJ && countJ > 0){
                        res = Math.max(res, countI - countJ);
                    }
                    
                    if(countI < countJ && remainJ > 0){
                        countI = 0;
                        countJ = 0;
                    }
                }
            }
        }
        
        return res;
    }
}

类似Maximum Subarray.