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Tested Python solution for LeetCode 532 with 16 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 532 # by problem number
lcpy gen -s k_diff_pairs_in_an_array # by problem name
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.lengthi != j|nums[i] - nums[j]| == k|val| denotes the absolute value of val.
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.
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).
Input: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
from collections import Counter
class Solution:
# Time: O(n)
# Space: O(n)
def find_pairs(self, nums: list[int], k: int) -> int:
counts = Counter(nums)
if k == 0:
return sum(1 for count in counts.values() if count > 1)
return sum(1 for value in counts if value + k in counts)
| Time | Space |
|---|---|
| O(n) | O(n) |