Problem Description

You have an integer array nums with 0-based indexing. You can rearrange the elements of this array in any order you want.

After rearranging the array, you need to calculate its prefix sum array. The prefix sum array prefix is defined such that prefix[i] equals the sum of all elements from index 0 to index i in the rearranged array.

Your goal is to find a rearrangement that maximizes the score, where the score is defined as the count of positive integers in the prefix sum array.

For example, if after rearranging you get nums = [2, -1, 3], then:

• prefix[0] = 2 (positive)
• prefix[1] = 2 + (-1) = 1 (positive)
• prefix[2] = 2 + (-1) + 3 = 4 (positive)

This would give a score of 3 since all three prefix sums are positive.

The task is to return the maximum possible score you can achieve by optimally rearranging the array elements.

Intuition

To maximize the number of positive prefix sums, we need to think about what makes a prefix sum positive. Each prefix sum is the cumulative total of elements up to that position. If we want as many positive prefix sums as possible, we should try to keep the running sum positive for as long as possible.

The key insight is that we should place larger positive numbers first. Why? Because when we add a large positive number early, it creates a "buffer" that can absorb negative numbers that come later while still keeping the prefix sum positive.

Consider this: if we have numbers like [5, -2, 3, -1], arranging them as [5, 3, -1, -2] (largest to smallest) gives us:

• First prefix: 5 (positive)
• Second prefix: 5 + 3 = 8 (positive)
• Third prefix: 8 + (-1) = 7 (positive)
• Fourth prefix: 7 + (-2) = 5 (positive)

All prefix sums remain positive! But if we had started with negative numbers or smaller positive numbers, we would lose the ability to maintain positive sums early on.

This leads us to a greedy strategy: sort the array in descending order. Start with the largest values to build up a strong positive sum, then add smaller values (including negatives) later. The accumulated positive sum from the beginning can help offset the negative values that come later.

We keep adding elements and calculating the running sum. Once the running sum becomes zero or negative, we know that from this point onward, all subsequent prefix sums will be non-positive (since we're adding smaller or more negative values). At that point, we can stop and return how many positive prefix sums we've counted so far.

Learn more about Greedy, Prefix Sum and Sorting patterns.

Solution Approach

The implementation follows a greedy approach with sorting:

1. Sort the array in descending order: We use nums.sort(reverse=True) to arrange elements from largest to smallest. This ensures we process the most positive (or least negative) elements first.

2. Iterate through the sorted array while maintaining a running sum: We initialize a variable s = 0 to track the prefix sum as we process each element.

3. For each element, update the prefix sum and check if it remains positive:

   for i, x in enumerate(nums):
       s += x
       if s <= 0:
           return i

   • We add the current element x to our running sum s
   • If at any point s becomes zero or negative, we immediately return the current index i
   • This index represents how many positive prefix sums we were able to achieve before the sum turned non-positive

4. Return the result:

   • If we complete the entire loop without the sum becoming non-positive, it means all prefix sums are positive, so we return len(nums)
   • Otherwise, we return the index where the sum first became non-positive

The algorithm has a time complexity of O(n log n) due to sorting, where n is the length of the array. The space complexity is O(1) if we don't count the space used by the sorting algorithm.

The key insight is that by processing elements in descending order, we maximize our chances of keeping the prefix sum positive for as many positions as possible. Once the sum drops to zero or below, we know that adding any remaining elements (which are smaller or more negative) won't help us achieve more positive prefix sums.

Example Walkthrough

Let's walk through the solution with a concrete example: nums = [2, -1, 0, 1, -3, 3, -3]

Step 1: Sort in descending order After sorting: nums = [3, 2, 1, 0, -1, -3, -3]

Step 2: Process elements and track prefix sums

Let's trace through each iteration:

Step 3: Return the result When we reach index 6, adding -3 makes our running sum negative (-1). At this point, we stop and return the current index, which is 6.

This means we can achieve a maximum score of 6 positive prefix sums by arranging the array as [3, 2, 1, 0, -1, -3, -3].

Notice how starting with the largest positive values (3, 2, 1) built up a "cushion" that allowed us to absorb the negative values (-1, -3) while keeping the prefix sum positive for as long as possible. The greedy strategy of sorting in descending order maximized our score.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log n), where n is the length of the array nums. This is dominated by the sorting operation nums.sort(reverse=True), which uses Python's Timsort algorithm with O(n × log n) complexity in the average and worst cases. The subsequent loop through the array takes O(n) time, but this is overshadowed by the sorting step.

The space complexity is O(log n). While Python's Timsort is an in-place sorting algorithm that modifies the original list, it still requires O(log n) auxiliary space for the recursion stack during the sorting process. The rest of the algorithm uses only O(1) additional space for variables s and i.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Termination Condition

The Issue: A common mistake is thinking that we should continue processing elements even after the prefix sum becomes zero or negative, hoping that future positive elements might make the sum positive again. This leads to incorrect implementations like:

def maxScore(self, nums: List[int]) -> int:
    nums.sort(reverse=True)
    prefix_sum = 0
    count = 0
  
    for value in nums:
        prefix_sum += value
        if prefix_sum > 0:  # Wrong: counting individual positive sums
            count += 1
  
    return count

Why It's Wrong: Once the prefix sum becomes zero or negative at position i, all subsequent prefix sums will also be non-positive because we've already used our largest elements. The remaining elements are smaller or more negative, so they cannot recover the sum to positive.

The Fix: Return immediately when the prefix sum becomes non-positive:

if prefix_sum <= 0:
    return index  # Return the count of positive prefix sums achieved

Pitfall 2: Incorrect Handling of All-Negative Arrays

The Issue: Forgetting to handle the edge case where all elements are negative. Some might incorrectly assume the answer should always be at least 1:

def maxScore(self, nums: List[int]) -> int:
    if all(x < 0 for x in nums):
        return 1  # Wrong assumption
    # ... rest of the code

Why It's Wrong: Even with all negative numbers, if we sort them in descending order (least negative first), the first element's prefix sum might still be positive if it's a negative number close to zero (e.g., -0.5 would still count as negative, but in integer arrays, the least negative would be -1).

The Fix: The original solution handles this correctly by checking if prefix_sum <= 0 which will return 0 if even the first (least negative) element makes the sum non-positive.

Pitfall 3: Off-by-One Error in Return Value

The Issue: Confusing the index with the count of positive prefix sums:

def maxScore(self, nums: List[int]) -> int:
    nums.sort(reverse=True)
    prefix_sum = 0
  
    for index, value in enumerate(nums):
        prefix_sum += value
        if prefix_sum <= 0:
            return index + 1  # Wrong: adds 1 unnecessarily
  
    return len(nums)

Why It's Wrong: When we encounter a non-positive sum at index i, we've successfully maintained positive sums for indices 0 through i-1, which is exactly i elements. Adding 1 would overcount.

The Fix: Return the index directly, as it represents the count of elements we successfully included while maintaining positive prefix sums:

if prefix_sum <= 0:
    return index  # Correct: index equals the count of positive prefix sums