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 <= 1050 <= nums[i] <= 1090 <= 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
}
}