How do you solve Maximum Score of a Split on LeetCode?

Maximum Score of a Split (LeetCode #3788) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Prefix sums help in quickly calculating cumulative values.

What is the time complexity of Maximum Score of a Split?

The optimal solution for Maximum Score of a Split runs in O(n) time and uses O(n) space. The complexity is linear because we traverse the array a few times to build prefix and suffix arrays.

What is the brute force solution for Maximum Score of a Split?

The brute force approach for Maximum Score of a Split is Brute Force, with O(n²) time complexity and O(1) space complexity. Calculate the score for each possible split index by iterating through the array. This method is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Score of a Split?

Maximum Score of a Split uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Prefix Sum, Suffix Array.

What are the interview tips for Maximum Score of a Split?

Explain your thought process clearly. Discuss time and space complexity upfront. Practice similar problems to build confidence.

What are common mistakes when solving Maximum Score of a Split?

Forgetting to handle edge cases like all negative numbers. Not considering the bounds of the split index correctly.

#3788

Maximum Score of a Split

Medium

ArrayPrefix SumPrefix SumSuffix Array

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Precompute prefix sums and suffix minimums to efficiently calculate scores in a single pass, reducing time complexity.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array where prefix[i] = sum of nums[0] to nums[i].
2Step 2: Create a suffix min array where suffix[i] = min of nums[i] to nums[n-1].
3Step 3: Iterate through valid split indices to compute scores using precomputed arrays and track the maximum score.

solution.py14 lines

1def maxScore(nums):
2    n = len(nums)
3    prefix = [0] * n
4    suffix = [0] * n
5    prefix[0] = nums[0]
6    for i in range(1, n):
7        prefix[i] = prefix[i - 1] + nums[i]
8    suffix[n - 1] = nums[n - 1]
9    for i in range(n - 2, -1, -1):
10        suffix[i] = min(suffix[i + 1], nums[i])
11    max_score = float('-inf')
12    for i in range(n - 1):
13        max_score = max(max_score, prefix[i] - suffix[i + 1])
14    return max_score

ℹ

Complexity note: The complexity is linear because we traverse the array a few times to build prefix and suffix arrays.

1Prefix sums help in quickly calculating cumulative values.
2Suffix minimums allow efficient retrieval of minimum values after a split.

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