Maximum Score After Splitting a String - Video Solutions

Problem Overview: You get a binary string s. Split it into two non-empty parts. The score equals the number of 0s in the left substring plus the number of 1s in the right substring. Your task is to find the split position that produces the maximum score.

The brute idea is simple: try every possible split and count zeros and ones on each side. But repeated counting leads to unnecessary work. The key insight is that the score can be computed incrementally if you track counts using prefix sum style preprocessing or by updating the score during a single scan.

Approach 1: Prefix Sum Approach (O(n) time, O(n) space)

Precompute counts so each split can be evaluated in constant time. Build a prefix array where prefixZero[i] stores the number of zeros from index 0 to i. Similarly track the total number of ones or compute a suffix-style value for ones on the right. For every split position i (between i and i+1), the score becomes prefixZero[i] + (totalOnes - onesPrefix[i]). Iterate through all valid split points and keep the maximum. This method uses the classic prefix sum technique and works well when you want clear precomputation and constant-time queries for each position in a string. Time complexity is O(n) because the string is scanned a few times, and space complexity is O(n) for the prefix arrays.

Approach 2: Single Pass Score Calculation (O(n) time, O(1) space)

The optimal approach avoids extra arrays. First count the total number of ones in the entire string. That represents the initial potential contribution of the right side. Then iterate from left to right, stopping before the last character so the right substring remains non-empty. When you encounter a 0, the left score increases by one. When you encounter a 1, the right side loses one because that 1 moves to the left partition. At each position compute currentScore = zerosLeft + onesRight and update the maximum. This technique effectively simulates all splits while maintaining running counts. Time complexity is O(n) and space complexity is O(1), making it the most efficient approach for large inputs.

Recommended for interviews: Start by explaining the idea of evaluating every split and how repeated counting would be inefficient. Then move to the prefix-sum reasoning to show how preprocessing removes redundant work. Finally present the single-pass method. Interviewers usually expect the O(n), O(1) solution because it demonstrates that you can transform prefix counting logic into an in-place running calculation.