[LeetCode October Challange] Day 3 - 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
• 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

---

Solution

Time complexity : O(n)
Space complexity : O(n)

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        unordered_map<int, int> ht;
        for (int n: nums)
            ++ht[n];
        
        int ans = 0;
        for (pair<int, int> item: ht) {
            if ((k==0 && item.second > 1) ||
                (k>0 && ht.count(item.first+k)) ) 
            {
                ++ans;
            }
        }
        
        return ans;
    }
};