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Formatted question description: https://leetcode.ca/all/2587.html

# 2587. Rearrange Array to Maximize Prefix Score

## Description

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

## Solutions

Solution 1: Greedy + Sorting

To maximize the number of positive integers in the prefix sum array, we need to make the elements in the prefix sum array as large as possible, that is, to add as many positive integers as possible. Therefore, we can sort the array $nums$ in descending order, then traverse the array, maintaining the prefix sum $s$. If $s \leq 0$, it means that there can be no more positive integers in the current position and the positions after it, so we can directly return the current position.

Otherwise, after the traversal, we return the length of the array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.

• 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;
}
}

• class Solution {
public:
int maxScore(vector<int>& nums) {
sort(nums.rbegin(), nums.rend());
long long s = 0;
int n = nums.size();
for (int i = 0; i < n; ++i) {
s += nums[i];
if (s <= 0) {
return i;
}
}
return n;
}
};

• 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)


• 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
}

• 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;
}


• 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
}
}