# 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
# i != j
# a <= b
# b - a == k

# 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 of unique 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).

# Example 4:

# Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
# Output: 2

# Example 5:

# Input: nums = [-1,-2,-3], k = 1
# Output: 2

# Constraints:

# 1 <= nums.length <= 10^4
# -10^7 <= nums[i] <= 10^7
# 0 <= k <= 107

class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        if k < 0: return 0
        result, lookup = set(), set()
        for num in nums:
            if num-k in lookup:
                result.add(num-k)
            if num+k in lookup:
                result.add(num)
            lookup.add(num)
        return len(result)

