K-diff Pairs in an Array

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).

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;
    }
};