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 (Java)

public class Solution {
    public int minimizeMax(int[] nums, int p) {
        Arrays.sort(nums);

        int left = 0, right = nums[nums.length - 1] - nums[0];

        while (left < right) {
            int mid = (left + right) / 2;
            if (can_form_pairs(nums, mid, p)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return left;
    }

    private boolean can_form_pairs(int[] nums, int mid, int p) {
        int count = 0;
        for (int i = 0; i < nums.length - 1 && count < p;) {
            if (nums[i+1] - nums[i] <= mid) {
                count++;
                i += 2;
            } else {
                i++;
            }
        }
        return count >= p;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).