532

532. K-diff Pairs in an Array (Medium)

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
• |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).

 

Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

Companies:
Goldman Sachs, Twitter, Amazon, Google

Related Topics:
Array, Hash Table, Two Pointers, Binary Search, Sorting

Similar Questions:

• Minimum Absolute Difference in BST (Easy)
• Count Number of Pairs With Absolute Difference K (Easy)
• Kth Smallest Product of Two Sorted Arrays (Hard)

Solution 1. Counting

Use an unordered_map<int, int> m to count the occurrence of each number.

For each number n, we increment the answer if one of the following is true:

• k == 0 and m[n] > 1.
• k != 0 and m[n + k] > 0.

// OJ: https://leetcode.com/problems/k-diff-pairs-in-an-array
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int findPairs(vector<int>& A, int k) {
        unordered_map<int, int> m;
        for (int n : A) m[n]++;
        int ans = 0;
        for (auto &[n, cnt] : m) {
            ans += k ? m.count(n - k) : cnt > 1;
        }
        return ans;
    }
};

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