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).
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
Solutions
🧠 Cpp
class Solution
{
public:
int findPairs(vector<int>& nums, int k)
{
size_t count = 0;
//O(n) solution, special case for k == 0
if(k == 0)
{
//num, number of entries
std::unordered_map<int, size_t> num_of_entries;
for(int num : nums)
num_of_entries[num]++;
for(auto &entrie : num_of_entries)
if(entrie.second > 1)
++count;
return count;
}
//O(n) solution for k > 0
//for O(1) lookups if element is present, O(N) extra space
std::unordered_set<int> lookup(begin(nums), end(nums));
//iterate set to check only 'unique' pairs
for(int num : lookup)
if(lookup.find(num+k) != end(lookup))
++count;
return count;
}
};