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
-
nums[i] - nums[j] == k
Notice that |val|
denotes the absolute value of val
.
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).
Constraints:
-
1 <= nums.length <= 104
-
-107 <= nums[i] <= 107
-
0 <= k <= 107
SOLUTION:
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
exists = {}
ctr = set()
for i, num in enumerate(nums):
exists[num] = exists.get(num, []) + [i]
for i, num in enumerate(nums):
for val in [num - k, num + k]:
if val in exists:
for j in exists[val]:
if j > i:
ctr.add(tuple(sorted([nums[i], nums[j]])))
return len(ctr)
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