532. K-diff Pairs in an Array 

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.

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

Solutions

Solution 1: Hash Table

Since $k$ is a fixed value, we can use a hash table $\textit{ans}$ to record the smaller value of the pairs, which allows us to determine the larger value. Finally, we return the size of $\textit{ans}$ as the answer.

We traverse the array $\textit{nums}$. For the current number $x$, we use a hash table $\textit{vis}$ to record all the numbers that have been traversed. If $x-k$ is in $\textit{vis}$, we add $x-k$ to $\textit{ans}$. If $x+k$ is in $\textit{vis}$, we add $x$ to $\textit{ans}$. Then, we add $x$ to $\textit{vis}$. Continue traversing the array $\textit{nums}$ until the end.

Finally, we return the size of $\textit{ans}$ as the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

Python3

class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        ans = set()
        vis = set()
        for x in nums:
            if x - k in vis:
                ans.add(x - k)
            if x + k in vis:
                ans.add(x)
            vis.add(x)
        return len(ans)

Java

class Solution {
    public int findPairs(int[] nums, int k) {
        Set<Integer> ans = new HashSet<>();
        Set<Integer> vis = new HashSet<>();
        for (int x : nums) {
            if (vis.contains(x - k)) {
                ans.add(x - k);
            }
            if (vis.contains(x + k)) {
                ans.add(x);
            }
            vis.add(x);
        }
        return ans.size();
    }
}

C++

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        unordered_set<int> ans, vis;
        for (int x : nums) {
            if (vis.count(x - k)) {
                ans.insert(x - k);
            }
            if (vis.count(x + k)) {
                ans.insert(x);
            }
            vis.insert(x);
        }
        return ans.size();
    }
};

Go

func findPairs(nums []int, k int) int {
	ans := make(map[int]struct{})
	vis := make(map[int]struct{})

	for _, x := range nums {
		if _, ok := vis[x-k]; ok {
			ans[x-k] = struct{}{}
		}
		if _, ok := vis[x+k]; ok {
			ans[x] = struct{}{}
		}
		vis[x] = struct{}{}
	}
	return len(ans)
}

TypeScript

function findPairs(nums: number[], k: number): number {
    const ans = new Set<number>();
    const vis = new Set<number>();
    for (const x of nums) {
        if (vis.has(x - k)) {
            ans.add(x - k);
        }
        if (vis.has(x + k)) {
            ans.add(x);
        }
        vis.add(x);
    }
    return ans.size;
}

Rust

use std::collections::HashSet;

impl Solution {
    pub fn find_pairs(nums: Vec<i32>, k: i32) -> i32 {
        let mut ans = HashSet::new();
        let mut vis = HashSet::new();

        for &x in &nums {
            if vis.contains(&(x - k)) {
                ans.insert(x - k);
            }
            if vis.contains(&(x + k)) {
                ans.insert(x);
            }
            vis.insert(x);
        }
        ans.len() as i32
    }
}