532. K-diff Pairs in an Array

1. Description

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.

2. Example

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

3. Constraints

• 1 <= nums.length <= $10^4$
• $-10^7$ <= nums[i] <= $10^7$
• 0 <= k <= $10^7$

4. Solutions

My Accepted Solution

n = i_nums.size()
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    // int findPairs(vector<int>& nums, int k)
    int findPairs(const vector<int>& i_nums, int k) {
        int pairsCount = 0;
        unordered_map<int, int> numTimes;
        for (auto num : i_nums) {
            ++numTimes[num];
        }

        for (auto num : i_nums) {
            if (k == 0 && numTimes[num] > 1 ||
                k != 0 && numTimes[num] > 0 && numTimes[num + k] != 0) {
                ++pairsCount;
                numTimes[num] = INT_MIN;
            }
        }

        return pairsCount;
    }
};

4.1 Two Pointers

n = i_nums.size()
Time complexity: O($nlog_2n$)
Space complexity: O(1)

// we sort the array firstly, then, we use two pointers to count pairs