K-diff Pairs in an Array
Question
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].
Tags
- Array
- Hash
Thought
Use hashmap to count the frequency of the number in the arrays and use that hashmap for counting the k-diff pairs.
Code
class Solution(object):
def findPairs(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
if k < 0:
return 0
counter = dict()
for num in nums:
if num not in counter:
counter[num] = 1
else:
counter[num] += 1
count = 0
for num in counter.keys():
current_num = num + k
if current_num not in counter:
continue
else:
if k == 0:
if counter[current_num] > 1:
count += 1
else:
count += 1
return count