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This approach uses a HashMap to track the frequency of each element in the array. For each unique element, we will check if there's another element that can form a k-diff pair with it. When k is zero, we need to check if there are duplicates present in the list.
Time Complexity: O(n) because we iterate over the array and then over the hashmap that is bounded by a constant size. Space Complexity: O(n) to store the frequency map.
1from collections import Counter
2
3def findPairs(nums, k):
4 count = 0
5 freq = Counter(nums)
6 for num in freq:
7 if k == 0:
8 if freq[num] > 1:
9 count += 1
10 else:
11 if num + k in freq:
12 count += 1
13 return countThis Python code uses a Counter to handle frequency counting. It correctly differentiates between the k=0 special case and positive k values, iterating over the elements in the frequency dictionary.
This approach involves sorting the array initially, then using two pointers to determine unique k-diff pairs. The array being sorted helps us efficiently reduce potential pair checks, and ensure pairs are considered only once.
Time Complexity: O(n log n) due to sorting. Space Complexity: O(1) if disregard input.
1functionJavaScript's solution first sorts the array, letting two pointers sieve the array for pairs with k differences. With special handling of duplicates, the unique k-diff pairs are effectively derived.