Question

532. K-diff Pairs in an Array

中文文档

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

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)

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();
  }
}

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();
  }
};

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)
}

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
  }
}