Question

561. Array Partition

中文文档

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

Solutions

Solution 1

class Solution:
  def arrayPairSum(self, nums: List[int]) -> int:
    return sum(sorted(nums)[::2])

class Solution {
  public int arrayPairSum(int[] nums) {
    Arrays.sort(nums);
    int ans = 0;
    for (int i = 0; i < nums.length; i += 2) {
      ans += nums[i];
    }
    return ans;
  }
}

class Solution {
public:
  int arrayPairSum(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int ans = 0;
    for (int i = 0; i < nums.size(); i += 2) ans += nums[i];
    return ans;
  }
};

func arrayPairSum(nums []int) int {
  sort.Ints(nums)
  ans := 0
  for i := 0; i < len(nums); i += 2 {
    ans += nums[i]
  }
  return ans
}

impl Solution {
  pub fn array_pair_sum(mut nums: Vec<i32>) -> i32 {
    nums.sort();
    let n = nums.len();
    let mut i = 0;
    let mut res = 0;
    while i < n {
      res += nums[i];
      i += 2;
    }
    res
  }
}

/**
 * @param {number[]} nums
 * @return {number}
 */
var arrayPairSum = function (nums) {
  nums.sort((a, b) => a - b);
  let ans = 0;
  for (let i = 0; i < nums.length; i += 2) {
    ans += nums[i];
  }
  return ans;
};