How do you solve Maximum Score After Splitting a String on LeetCode?

Maximum Score After Splitting a String (LeetCode #1422) 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: The score is maximized by balancing zeros on the left and ones on the right.

What is the time complexity of Maximum Score After Splitting a String?

The optimal solution for Maximum Score After Splitting a String runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the string a constant number of times, making it efficient.

What is the brute force solution for Maximum Score After Splitting a String?

The brute force approach for Maximum Score After Splitting a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to split the string into two non-empty parts. For each split, we count the zeros in the left part and the ones in the right part to calculate the score. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Score After Splitting a String?

Maximum Score After Splitting a String uses the following concepts: String, Prefix Sum. The recommended patterns to study are: Prefix Sum, Two Pointers.

What are the interview tips for Maximum Score After Splitting a String?

Always start with a brute force solution to understand the problem. Look for patterns in the problem that can lead to optimizations. Practice explaining your thought process clearly during the interview.

What are common mistakes when solving Maximum Score After Splitting a String?

Not considering all possible split points, especially the last character. Confusing the counts of zeros and ones when calculating scores.

#1422

Maximum Score After Splitting a String

Easy

StringPrefix SumPrefix SumTwo Pointers

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)

The optimal solution uses a single pass to calculate the score efficiently. We maintain a count of zeros in the left substring and derive the count of ones in the right substring using the total count of ones.

⚙️

Algorithm

4 steps

1Step 1: Count the total number of ones in the string.
2Step 2: Initialize a variable to keep track of the number of zeros in the left substring and the maximum score.
3Step 3: Iterate through the string up to the second last character, updating the count of zeros and calculating the score using the total ones minus the ones in the left substring.
4Step 4: Update the maximum score accordingly.

solution.py12 lines

1# Full working Python code
2
3def maxScore(s):
4    total_ones = s.count('1')
5    left_zeros = 0
6    max_score = 0
7    for i in range(len(s) - 1):
8        if s[i] == '0':
9            left_zeros += 1
10        score = left_zeros + (total_ones - (s[:i + 1].count('1')))
11        max_score = max(max_score, score)
12    return max_score

ℹ

Complexity note: This complexity is linear because we only traverse the string a constant number of times, making it efficient.

1The score is maximized by balancing zeros on the left and ones on the right.
2Precomputing the total number of ones allows for efficient score calculation.

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