LeetCode in Kotlin

2006. Count Number of Pairs With Absolute Difference K

Easy

Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.

The value of |x| is defined as:

• x if x >= 0.
• -x if x < 0.

Example 1:

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

Output: 4

Explanation: The pairs with an absolute difference of 1 are:

• [1,2,2,1]

• [1,2,2,1]

• [1,2,2,1]

• [1,2,2,1]

Example 2:

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

Output: 0

Explanation: There are no pairs with an absolute difference of 3.

Example 3:

Input: nums = [3,2,1,5,4], k = 2

Output: 3

Explanation: The pairs with an absolute difference of 2 are:

• [3,2,1,5,4]

• [3,2,1,5,4]

• [3,2,1,5,4]

Constraints:

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100
• 1 <= k <= 99

Solution

import kotlin.math.abs

class Solution {
    fun countKDifference(nums: IntArray, k: Int): Int {
        var pairs = 0
        for (i in 0 until nums.size - 1) {
            for (j in i + 1 until nums.size) {
                if (abs(nums[i] - nums[j]) == k) {
                    pairs++
                }
            }
        }
        return pairs
    }
}

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