561. Array Partition

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

• 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;
}
};

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


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

• /**
* @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;
};


• 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
}
}