How do you solve Equal Score Substrings on LeetCode?

Equal Score Substrings (LeetCode #3707) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Prefix sums can optimize score calculations.

What is the time complexity of Equal Score Substrings?

The optimal solution for Equal Score Substrings runs in O(n) time and uses O(1) space. We calculate the total score in O(n) and use a single pass to check splits, leading to O(n) overall.

What is the brute force solution for Equal Score Substrings?

The brute force approach for Equal Score Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible split of the string and calculate the scores for both substrings. If any split yields equal scores, we return true. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Equal Score Substrings?

Equal Score Substrings uses the following concepts: String, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Equal Score Substrings?

Clarify the problem constraints. Discuss brute force before jumping to optimizations. Explain your thought process clearly.

What are common mistakes when solving Equal Score Substrings?

Not considering all possible splits. Miscalculating character scores.

#3707

Equal Score Substrings

Easy

StringPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can use a prefix sum to calculate scores efficiently. By maintaining a running total, we can quickly determine if a split yields equal scores.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total score of the string.
2Step 2: Iterate through the string, maintaining a running score for the left substring.
3Step 3: At each index, check if twice the left score equals the total score minus the score of the character at the current index.

solution.py8 lines

1def equal_score_substrings(s):
2    total_score = sum(ord(c) - ord('a') + 1 for c in s)
3    left_score = 0
4    for i in range(len(s) - 1):
5        left_score += ord(s[i]) - ord('a') + 1
6        if 2 * left_score == total_score - (ord(s[i + 1]) - ord('a') + 1):
7            return True
8    return False

ℹ

Complexity note: We calculate the total score in O(n) and use a single pass to check splits, leading to O(n) overall.

1Prefix sums can optimize score calculations.
2Equal score condition can be checked with simple arithmetic.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.