LeetCode 0561. Array Partition Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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: Sorting

For a pair of numbers (a, b), we can assume a ≤ b, then min(a, b) = a. In order to make the sum as large as possible, the b we choose should be as close to a as possible, so as to retain a larger number.

Therefore, we can sort the array nums, then divide every two adjacent numbers into a group, and add the first number of each group.

The time complexity is O(n × log n), and the space complexity is O(log n). Where n is the length of the array nums.

PythonJavaC++GoRustJavaScript

class Solution: def arrayPairSum(self, nums: List[int]) -> int: nums.sort() return sum(nums[::2])(code-box)

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; } }(code-box)

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; } };(code-box)

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

impl Solution { pub fn array_pair_sum(mut nums: Vec<i32>) -> i32 { nums.sort(); nums.iter().step_by(2).sum() } }(code-box)

/** * @param {number[]} nums * @return {number} */ var arrayPairSum = function (nums) { nums.sort((a, b) => a - b); return nums.reduce((acc, cur, i) => (i % 2 === 0 ? acc + cur : acc), 0); };(code-box)