K-diff pairs in an array

Time: O(N); Space: O(N); easy

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array.

Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is 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).

Notes:

  • The pairs (i, j) and (j, i) count as the same pair.

  • The length of the array won’t exceed 10,000.

  • All the integers in the given input belong to the range: [-1e7, 1e7].

[2]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(N)
    """
    def findPairs(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: 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)

[3]:

s = Solution1()

nums = [3, 1, 4, 1, 5]
k = 2
assert s.findPairs(nums, k) == 2

nums = [1, 2, 3, 4, 5]
k = 1
assert s.findPairs(nums, k) == 4

nums = [1, 3, 1, 5, 4]
k = 0
assert s.findPairs(nums, k) == 1