Welcome to Subscribe On Youtube

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

• Java
• C++
• Python
• Go
• RenderScript

• 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();
      }
  };
  

• 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)
  
  

• 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
      }
  }
  
  

All Problems

All Solutions