1422. 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.

Approach 01:

#include <string>
#include <algorithm>
using namespace std;

class Solution {
public:
    int maxScore(string binaryString) {
        int stringLength = binaryString.length();
        int totalOnesInString = 0;  // Total number of '1's in the string
        
        // Count the total number of '1's in the string
        for (char character : binaryString) {
            if (character == '1') {
                ++totalOnesInString;
            }
        }

        int leftZeroCount = 0, rightOneCount = totalOnesInString, maximumScore = 0;
        
        // Iterate up to the second last character to ensure both parts are non-empty
        for (int index = 0; index < stringLength - 1; ++index) {
            if (binaryString[index] == '0') {
                ++leftZeroCount;
            } else {
                --rightOneCount;
            }

            // Calculate the score and update the maximum score
            maximumScore = max(maximumScore, leftZeroCount + rightOneCount);
        }

        return maximumScore;
    }
};

class Solution:
    def maxScore(self, binaryString: str) -> int:
        stringLength = len(binaryString)
        totalOnesInString = binaryString.count('1')  # Total number of '1's in the string
        
        leftZeroCount = 0
        rightOneCount = totalOnesInString
        maximumScore = 0
        
        # Iterate up to the second last character to ensure both parts are non-empty
        for index in range(stringLength - 1):
            if binaryString[index] == '0':
                leftZeroCount += 1
            else:
                rightOneCount -= 1

            # Calculate the score and update the maximum score
            maximumScore = max(maximumScore, leftZeroCount + rightOneCount)
        
        return maximumScore

Time Complexity:

• Counting ‘1’s:
  • We iterate through the entire string to count the total number of ‘1’s.
  • This operation takes \( O(n) \), where \( n \) is the length of the string.
• Iterating to Calculate Scores:
  • We traverse the string again, up to the second-last character, to calculate scores.
  • This also takes \( O(n) \) time.
• Overall Time Complexity:

  \( O(n) \), as the operations to count ‘1’s and calculate scores dominate the runtime.

Space Complexity:

• Auxiliary Variables:
  • We use a constant amount of additional space for variables such as leftZeroCount, rightOneCount, and maximumScore.
• Input Storage:
  • The input string is provided as an argument and does not require extra space to copy or process.
• Overall Space Complexity:

  \( O(1) \), as the algorithm only uses a fixed amount of space regardless of the input size.