2616. Minimize the Maximum Difference of Pairs

Problem

You are given a 0-indexed integer array nums and an integer p. Find p pairs of indices of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure no index appears more than once amongst the p pairs.

Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x.

Return **the *minimum* maximum difference among all **p *pairs.* We define the maximum of an empty set to be zero.

  Example 1:

Input: nums = [10,1,2,7,1,3], p = 2
Output: 1
Explanation: The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. 
The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.

Example 2:

Input: nums = [4,2,1,2], p = 1
Output: 0
Explanation: Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.

  Constraints:

• 1 <= nums.length <= 105

• 0 <= nums[i] <= 109

• 0 <= p <= (nums.length)/2

Solution

/**
 * @param {number[]} nums
 * @param {number} p
 * @return {number}
 */
var minimizeMax = function(nums, p) {
    nums.sort((a, b) => a - b);
    var left = 0;
    var right = nums[nums.length - 1] - nums[0];
    while (left < right) {
        var mid = left + Math.floor((right - left) / 2);
        if (count(nums, mid) >= p) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
};

var count = function(nums, n) {
    var num = 0;
    for (var i = 0; i < nums.length - 1; i++) {
        if (nums[i + 1] - nums[i] <= n) {
            i++;
            num++;
        }
    }
    return num;
}

Explain:

nope.

Complexity:

• Time complexity : O(n * log(max - min)).
• Space complexity : O(1).