LeetCode 532 LintCode 1187 K-diff Pairs in An Array

Table of contents

Description

Question Links: LeetCode 532, LintCode 1187

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).

Constraints:

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

Idea

We could count the values in the array and then iterate through the counter.

1. If k==0, we count if this element count is >= 2. Because we would only consider the unique pairs, if the count is <= 1, there is no such pair; if the count is >= 2, we count the one pair of v and v.
2. If k>0, we look for v+k in the counter.

Complexity: Time $O(n)$, Space $O(n)$.

C++

// leet 532, 3 ms, 17.22 mb
class Solution {
public:
    int findPairs(vector<int> &nums, int k) {
        int res = 0;
        unordered_map<int, int> cnt;
        for (auto &n: nums) cnt[n] += 1;
        for (auto e: cnt) res += k > 0 && cnt.count(e.first + k) || k == 0 && e.second > 1;
        return res;
    }
};

Python

class Solution:
    """5 ms, 19 mb"""

    def findPairs(self, nums: list[int], k: int) -> int:
        cnt = Counter(nums)
        return sum([k > 0 and c + k in cnt or k == 0 and cnt[c] > 1 for c in cnt])