[Leetcode]561. Array Partition I

김지원·2022년 4월 8일
0

📄 Description

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

Constraints:

  • 1 <= n <= 104
  • nums.length == 2 * n
  • -104 <= nums[i] <= 104

💻 My Submission

class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        nums.sort()
        
        return sum([x for i, x in enumerate(nums) if i%2==0])

💊 Other solution

  • using slicing the nums list! (슬라이싱 사용)
class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
		return sum(sorted(nums)[::2])

References

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