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

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

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

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

Go Code
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func findPairs(nums []int, k int) int {
	vis := map[int]bool{}
	ans := map[int]bool{}
	for _, v := range nums {
		if vis[v-k] {
			ans[v-k] = true
		}
		if vis[v+k] {
			ans[v] = true
		}
		vis[v] = true
	}
	return len(ans)
}

Rust Code
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impl Solution {
    pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
        nums.sort();
        let n = nums.len();
        let mut res = 0;
        let mut left = 0;
        let mut right = 1;
        while right < n {
            let num = i32::abs(nums[left] - nums[right]);
            if num == k {
                res += 1;
            }
            if num <= k {
                right += 1;
                while right < n && nums[right - 1] == nums[right] {
                    right += 1;
                }
            } else {
                left += 1;
                while left < right && nums[left - 1] == nums[left] {
                    left += 1;
                }
                if left == right {
                    right += 1;
                }
            }
        }
        res
    }
}