2616. Minimize the Maximum Difference of Pairs

Description

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

 

Solutions

Solution: Binary Search + Greedy

• Time complexity: O(nlogn+nlog(maxDiff))
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} p
 * @return {number}
 */
const minimizeMax = function (nums, p) {
  const n = nums.length;

  nums.sort((a, b) => a - b);

  let left = 0;
  let right = nums[n - 1] - nums[0];

  const pairsCount = diff => {
    let index = 1;
    let count = 0;

    while (index < n) {
      const current = nums[index] - nums[index - 1];

      if (current <= diff) {
        count += 1;
        index += 2;
      } else {
        index += 1;
      }
    }

    return count;
  };

  while (left < right) {
    const mid = Math.floor((left + right) / 2);

    pairsCount(mid) >= p ? (right = mid) : (left = mid + 1);
  }

  return left;
};