Oct 3, 2021 - 1 ' read

Count Number of Pairs With Absolute Difference K

Problem Statement - link #

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.

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]  (index 0,1)
- [1,2,2,1]  (index 0,2)
- [1,2,2,1]  (index 1,3)
- [1,2,2,1]  (index 2,3)

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]  (index 0,2)
- [3,2,1,5,4]  (index 0,3)
- [3,2,1,5,4]  (index 1,4)

Constraints #

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

Solution #

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int count = 0;
        for(int i = 0; i < nums.size()-1 ; i++){
            for(int j = i+1; j < nums.size() ; j++){
                if( (nums[j] - nums[i] == k ) || (nums[i] - nums[j] == k ) ){
                    count++;
                }
            }
        }
        return count;
    }
};