LeetCode 3707: Equal Score Substrings ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2025-10-11

Submissions: Wrong Answer → Accepted

Approach
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Compute the total score, then accumulate the left score from the left; the right side is total - left (Prefix Sum Template 2). Both sides must be non-empty, so stop before the last character.

My Attempts
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Attempt 1 used ord(char) as the score. ord is the alphabet position plus 96, so sides of different lengths gain a different 96 × length. For "bac", "ba" scores 2 + 1 = 3 and "c" scores 3, so the answer is True; with ord it is 195 versus 99, returning False. Using ord(char) - ord("a") + 1 passed.

Solution
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class Solution:
    def scoreBalance(self, s: str) -> bool:
        scores = [ord(c) - ord("a") + 1 for c in s]
        total, left = sum(scores), 0
        for x in scores[:-1]:
            left += x
            if left * 2 == total:
                return True
        return False

Complexity
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Time O(n), space O(n).

Mistakes to Avoid
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The score is the alphabet position 1 to 26: subtract ord("a") and add 1.
The right side cannot be empty, so the last character is not a split point.

Interview Explanation
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I compute the whole string’s score, then accumulate from the left; when the left is exactly half the total, a split works. Scores are alphabet positions, ord(c) - ord('a') + 1. Time O(n).