DEV Community

Abhishek Chaudhary

Posted on

K-diff Pairs in an Array

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: