2006. Count Number of Pairs With Absolute Difference K.py

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        ans = []
        for i in range(len(nums)):
            ans+=[ [nums[i],nums[j]] for j in range(i+1, len(nums)) if abs(nums[i] - nums[j]) == k]
        return len(ans)
        
        
################# OR ############################

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        ans = []
        for i in range(len(nums)):
            for j in range(i,len(nums)):
                if abs(nums[i] - nums[j]) == k:
                    ans.append([nums[i], nums[j]])
        return len(ans)


'''

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


'''