LeetCode in Kotlin

2616. Minimize the Maximum Difference of Pairs

Medium

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

@Suppress("NAME_SHADOWING")
class Solution {
    private fun ispossible(nums: IntArray, p: Int, diff: Int): Boolean {
        var p = p
        val n = nums.size
        var i = 1
        while (i < n) {
            if (nums[i] - nums[i - 1] <= diff) {
                p--
                i++
            }
            i++
        }
        return p <= 0
    }

    fun minimizeMax(nums: IntArray, p: Int): Int {
        val n = nums.size
        nums.sort()
        var left = 0
        var right = nums[n - 1] - nums[0]
        var ans = right
        while (left <= right) {
            val mid = left + (right - left) / 2
            if (ispossible(nums, p, mid)) {
                ans = mid
                right = mid - 1
            } else left = mid + 1
        }
        return ans
    }
}

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