Maximum Score After Splitting a String

Leet Code Link - https://leetcode.com/problems/maximum-score-after-splitting-a-string/description/

Complexity - Easy

Description

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.

Solution

We can count ones and zeros in the first iteration and then from second modify the count

class Solution {
    public int maxScore(String s) {
        int leftSum =0; int rightSum = 0;  int maxsum = 0;
        int j=1;
        for(int i=0;i<s.length()-1;i++){
            if(i==0){
                if(s.charAt(0)=='0'){
                    leftSum++;
                }
                while(j<s.length()){
                    if(s.charAt(j)=='1'){
                        rightSum++;
                    }
                    j++;
                }
            }else{
                if(s.charAt(i)=='0'){
                    leftSum++;
                }else if(s.charAt(i)=='1'){
                    rightSum--;
                }
            }

                maxsum=Math.max(maxsum,leftSum+rightSum);
        }
        return maxsum;
    }
}

Let us see more optimized 1 pass solution

class Solution {
    public int maxScore(String s) {
        int ones = 0;
        int zeros = 0;
        int best = Integer.MIN_VALUE;
        
        for (int i = 0; i < s.length() - 1; i++) {
            if (s.charAt(i) == '1') {
                ones++;
            } else {
                zeros++;
            }
            
            best = Math.max(best, zeros - ones);
        }
        
        if (s.charAt(s.length() - 1) == '1') {
            ones++;
        }
        
        return best + ones;
    }
}

Given n as the length of nums,

Time complexity: O(n)

We make one pass over nums, performing O(1) work at each iteration.

Space complexity: O(1)

We aren't using any extra space other than a few integers.