LeetCode #2587 — MEDIUM

Rearrange Array to Maximize Prefix Score

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed integer array nums. You can rearrange the elements of nums to any order (including the given order).

Let prefix be the array containing the prefix sums of nums after rearranging it. In other words, prefix[i] is the sum of the elements from 0 to i in nums after rearranging it. The score of nums is the number of positive integers in the array prefix.

Return the maximum score you can achieve.

Example 1:

Input: nums = [2,-1,0,1,-3,3,-3]
Output: 6
Explanation: We can rearrange the array into nums = [2,3,1,-1,-3,0,-3].
prefix = [2,5,6,5,2,2,-1], so the score is 6.
It can be shown that 6 is the maximum score we can obtain.

Example 2:

Input: nums = [-2,-3,0]
Output: 0
Explanation: Any rearrangement of the array will result in a score of 0.

Constraints:

1 <= nums.length <= 10⁵
-10⁶ <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. You can rearrange the elements of nums to any order (including the given order). Let prefix be the array containing the prefix sums of nums after rearranging it. In other words, prefix[i] is the sum of the elements from 0 to i in nums after rearranging it. The score of nums is the number of positive integers in the array prefix. Return the maximum score you can achieve.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[2,-1,0,1,-3,3,-3]

Example 2

[-2,-3,0]

Core Insight

What unlocks the optimal approach

The best order of the array is in decreasing order.
Sort the array in decreasing order and count the number of positive values in the prefix sum array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2587: Rearrange Array to Maximize Prefix Score
class Solution {
    public int maxScore(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        long s = 0;
        for (int i = 0; i < n; ++i) {
            s += nums[n - i - 1];
            if (s <= 0) {
                return i;
            }
        }
        return n;
    }
}

// Accepted solution for LeetCode #2587: Rearrange Array to Maximize Prefix Score
func maxScore(nums []int) int {
	sort.Ints(nums)
	n := len(nums)
	s := 0
	for i := range nums {
		s += nums[n-i-1]
		if s <= 0 {
			return i
		}
	}
	return n
}

# Accepted solution for LeetCode #2587: Rearrange Array to Maximize Prefix Score
class Solution:
    def maxScore(self, nums: List[int]) -> int:
        nums.sort(reverse=True)
        s = 0
        for i, x in enumerate(nums):
            s += x
            if s <= 0:
                return i
        return len(nums)

// Accepted solution for LeetCode #2587: Rearrange Array to Maximize Prefix Score
impl Solution {
    pub fn max_score(mut nums: Vec<i32>) -> i32 {
        nums.sort_by(|a, b| b.cmp(a));
        let mut s: i64 = 0;
        for (i, &x) in nums.iter().enumerate() {
            s += x as i64;
            if s <= 0 {
                return i as i32;
            }
        }
        nums.len() as i32
    }
}

// Accepted solution for LeetCode #2587: Rearrange Array to Maximize Prefix Score
function maxScore(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let s = 0;
    for (let i = 0; i < n; ++i) {
        s += nums[n - i - 1];
        if (s <= 0) {
            return i;
        }
    }
    return n;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.