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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
-
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; }