Robert BaindourovNodeJS, ReactJS, ElectronJS - Web Software Architect

Algorithm Intuition and Strategy

A naive brute-force solution would involve computing the left zeros and right ones for every possible split point independently. This nested approach would result in $O(N^2)$ time complexity, which is computationally expensive for large strings. However, we can heavily optimize this process to achieve an $O(N)$ single-pass linear time complexity by leveraging a preliminary counting step.

The trick is to pre-calculate the total number of ones present in the entire string. As we iterate through the characters from left to right to test each potential split point, we dynamically adjust our counts. If we encounter a '0', our left zero count increases. If we encounter a '1', our right one count decreases, because that '1' has effectively moved from the right substring into the left substring. This dynamic tallying allows us to continuously track the maximum score efficiently.

Optimized C++ Implementation

class Solution {
public:
    int maxScore(string s) {
        int ones = count(s.begin(), s.end(), '1'); // Count total 1's initially
        int zeros = 0, result = 0;

        for (int i = 0; i < s.size() - 1; i++) { // Leave at least one character for the right part
            if (s[i] == '1') 
                ones--; // A '1' shifts from right to left
            else 
                zeros++; // A '0' is added to the left part

            result = max(result, zeros + ones); // Update maximum score found so far
        }

        return result;
    }
};

Computational Complexity Analysis

• Time Complexity: $O(N)$. We make one initial pass through the string to count the ones, followed by a second pass to evaluate the split points. The operations inside the loop run in constant time.
• Space Complexity: $O(1)$. We strictly use a few integer variables (ones, zeros, result) for tracking counts, requiring no auxiliary data structures that scale with the input size.

Frequently Asked Questions (FAQ)

Why do we stop the loop at s.size() - 1?

The problem strictly requires that both the left and right substrings must be non-empty. By stopping the iteration at s.size() - 1, we guarantee that at least one character remains in the right substring, fulfilling the problem's constraints.

Can this algorithm handle strings with only zeros or only ones?

Yes, the logic holds perfectly. If the string contains only zeros, the ones variable starts at 0 and remains 0, while zeros increments, eventually giving the correct score. A similar resilient behavior applies when the string contains exclusively ones.

Is there a way to solve this in a single pass without pre-counting?

Yes, you can rewrite the score equation to optimize it into a true one-pass algorithm. Since the score is zeros_left + ones_right, and ones_right = total_ones - ones_left, the score becomes zeros_left - ones_left + total_ones. By maximizing zeros_left - ones_left during a single pass and adding total_ones at the end, you avoid the pre-counting loop, though the complexity remains $O(N)$.