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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.
1#include <vector>
#include <algorithm>
int findPairs(std::vector<int>& nums, int k) {
std::sort(nums.begin(), nums.end());
int count = 0, left = 0, right = 0;
while (right < nums.size()) {
if (right == left || nums[right] - nums[left] < k) {
++right;
} else if (nums[right] - nums[left] > k) {
++left;
} else { // nums[right] - nums[left] == k
++count;
++left;
while (right < nums.size() && nums[right] == nums[right - 1]) {
++right;
}
}
}
return count;
}The array is sorted initially, after which two pointers determine the valid pairs by comparing the difference to k. If a valid pair is found, the left pointer moves forward. Duplicate elements are effectively managed.