532. K-diff Pairs in an Array
Medium
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).
Example 4:
Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3 Output: 2
Example 5:
Input: nums = [-1,-2,-3], k = 1 Output: 2
Constraints:
1 <= nums.length <= 104
-107 <= nums[i] <= 107
0 <= k <= 107
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | class Solution { public int findPairs(int[] nums, int k) { Arrays.sort(nums); int n = nums.length; int cnt = 0; Set<Integer> set = new HashSet<>(); for(int i=0; i<n; ++i){ if(i>0 && nums[i]==nums[i-1]){ if(k>0) continue; if(k==0 && i>1 && nums[i-1]==nums[i-2]) continue; } int v = nums[i]; if(set.contains(v-k)) cnt += 1; set.add(v); } return cnt; } } |
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