k-diff-pairs-in-an-array.py

# Time:  O(n)
# Space: O(n)

# Total Accepted: 5671
# Total Submissions: 20941
# Difficulty: Easy
# Contributors: murali.kf370
# 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: [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:[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: [1, 3, 1, 5, 4], k = 0
# Output: 1
# Explanation: There is one 0-diff pair in the array, (1, 1).
# Note:
# 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].

class Solution(object):
    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)