0532. 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
• -10^7 <= nums[i] <= 10^7
• 0 <= k <= 10^7

Analysis

Edge cases: 1. k == 0, in this case, we need to return the pair that they two elements are equal to each other. 2. k < 0, should return 0. 3. handling duplicates pairs: (i, j) = (j, i)

Use a unordered_map<int, int> to keep track of the cnt of each appeared elements.

• if k == 0, we just need to filter out all the elements that have the count greater than 1.
• else check the complement (curr element + k) if exist in the map. don't need to check element - k.

Time: $O(n)$ Space: $O(n)$ for the unordered_map

Code

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        int res = 0;
        unordered_map<int, int> m;
        for (int num : nums) ++m[num];
        for (auto a : m) {
            if (k == 0 && a.second > 1) ++res; // multiple duplicate
            if (k > 0 && m.count(a.first + k)) ++res;
        }
        return res;
    }
};

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Last update: April 1, 2022