Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

• 0 <= i < j < nums.length
• |nums[i] - nums[j]| == k

Notice that |val| denotes the absolute value of val.

Example 1:

Input: nums = [3,1,4,1,5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number ofunique pairs.

Example 2:

Input: nums = [1,2,3,4,5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Constraints:

• 1 <= nums.length <= 104
• 107 <= nums[i] <= 107
• 0 <= k <= 107

Key Thoughts: Whenever key+k is in the set, the pair amount increases by 1. Don’t need to delete a whole 等差数列.

from collections import Counter
class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        # Use Counter 
        # O(n) , O(n)
        
        count = 0
        
        diction = Counter(nums)
        
        if k == 0: # then is value >1 means have duplicate then, count += 1
            for du in diction.values():
                if du > 1:
                    count += 1
            # print(diction)
        else:
            for key in diction.keys():
                if key+k in diction:
                    count += 1
        return count