Leetcode problem 84 - Array Partition

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.

/**
 * @param {number[]} nums
 * @return {number}
 */
var arrayPairSum = function (nums) {
  // Initialize a variable to store the sum of minimum values in pairs
  let result = 0;

  // Sort the 'nums' array in ascending order
  nums.sort(function (a, b) {
    return a - b;
  });

  // Iterate over the sorted array by skipping one element at a time (every 2 elements)
  for (let i = 0; i < nums.length; i += 2) {
    // Add the current minimum value in the pair to the result
    result += nums[i];
  }

  // Return the accumulated result
  return result;
};

The time complexity of this function is O(nlogn) because of the sorting algorithm used. The nums.sort function typically has a time complexity of O(n log n), where n is the length of the array nums. The space complexity is O(1) because no extra space is used.