Solution added.

30 Days of October Challange/Week 1/3. K-diff Pairs in an Array/solution.py

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+"""
+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
+a <= b
+b - a == k
+
+
+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).
+Example 4:
+
+Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
+Output: 2
+Example 5:
+
+Input: nums = [-1,-2,-3], k = 1
+Output: 2
+
+
+Constraints:
+
+1 <= nums.length <= 104
+-107 <= nums[i] <= 107
+0 <= k <= 107
+"""
+
+
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        hmap, result = {}, set()
+
+        for i, num in enumerate(nums):
+            hmap[num] = i
+
+        for j, num in enumerate(nums):
+            if num - k in hmap and (num, num - k) not in result and hmap[num - k] != j:
+                result.add((num - k, num))
+        return len(result)