2006. Count Number of Pairs With Absolute Difference K

Description

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.

Constraints

• 1 <= nums.length <= 200

• 1 <= nums[i] <= 100

• 1 <= k <= 99

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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]

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int countKDifference(int[] nums, int k) {
        int n = nums.length, count = 0;
        
        for(int i = 0; i < n; i++) {
            for(int j = i+1; j < n; j++) {
                if(Math.abs(nums[i]-nums[j]) == k) {
                    count++;
                }
            }
        }
        
        return count;
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */
 
 class Solution {
    public int countKDifference(int[] nums, int k) {
        if(nums == null || nums.length < 1) {
            return 0;
        }
        int[] counter = new int[101];
        for(int num: nums) {
            counter[num]++;
        }
        
        int count = 0;
        for(int i = k+1; i < counter.length; i++) {
            count += (counter[i-k] * counter[i]);
        }
        
        return count;
    }
}

Follow up