1422. Maximum Score After Splitting a String

https://leetcode.com/problems/maximum-score-after-splitting-a-string/

Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).

The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.

Example 1:

Input: s = "011101"
Output: 5 
Explanation: 
All possible ways of splitting s into two non-empty substrings are:
left = "0" and right = "11101", score = 1 + 4 = 5 
left = "01" and right = "1101", score = 1 + 3 = 4 
left = "011" and right = "101", score = 1 + 2 = 3 
left = "0111" and right = "01", score = 1 + 1 = 2 
left = "01110" and right = "1", score = 2 + 1 = 3

Example 2:

Input: s = "00111"
Output: 5
Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5

Example 3:

Input: s = "1111"
Output: 3

Constraints:

• 2 <= s.length <= 500

• The string s consists of characters '0' and '1' only.

由01组成的数组问左右划分后使得左边的count(l_0)和右边的count(l_1)的和最大（要求左右至少有一个元素），这个argmax(sum(count(l_0) + count(l_1)))是多少。直观的解法是前后各遍历一遍记录下以i点分界前面的0和后面的1的个数。实际只需要从头遍历一遍，以i为左边最后一个元素，记录当前遇到的0和1的个数，所求为argmax(count(l_0) + count(r_1)) = argmax(count(l_0) - count(l_1) + count(total_1)) ，因此遍历时再比较当前是否argmax(count(l_0) - count(l_1))。

class Solution:
    def maxScore(self, s: str) -> int:
        zs = os = 0
        ldiff = -float('inf')
        for i, c in enumerate(s):
            if c == '0':
                zs += 1
            else:
                os += 1
            if i != len(s) - 1:
                ldiff = max(ldiff, zs - os)
        return ldiff + os