LeetCode in Kotlin

719. Find K-th Smallest Pair Distance

Hard

The distance of a pair of integers a and b is defined as the absolute difference between a and b.

Given an integer array nums and an integer k, return the kth smallest distance among all the pairs nums[i] and nums[j] where 0 <= i < j < nums.length.

Example 1:

Input: nums = [1,3,1], k = 1

Output: 0

Explanation: Here are all the pairs:

(1,3) -> 2

(1,1) -> 0

(3,1) -> 2

Then the 1^st smallest distance pair is (1,1), and its distance is 0.

Example 2:

Input: nums = [1,1,1], k = 2

Output: 0

Example 3:

Input: nums = [1,6,1], k = 3

Output: 5

Constraints:

• n == nums.length
• 2 <= n <= 104
• 0 <= nums[i] <= 106
• 1 <= k <= n * (n - 1) / 2

Solution

class Solution {
    fun smallestDistancePair(nums: IntArray, k: Int): Int {
        nums.sort()
        val length = nums.size
        val maxDiff = nums[length - 1] - nums[0]
        var start = 0
        var end = maxDiff
        while (start < end) {
            val mid = start + (end - start) / 2
            if (isPair(nums, mid, k)) {
                end = mid
            } else {
                start = mid + 1
            }
        }
        return start
    }

    private fun isPair(nums: IntArray, mid: Int, k: Int): Boolean {
        var count = 0
        var i = 0
        for (j in 1 until nums.size) {
            while (nums[j] - nums[i] > mid) {
                i++
            }
            count += j - i
        }
        return count >= k
    }
}

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