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
a <= b
b - a == k
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
two pointers, time = O(n), space = O(n)
class Solution { public int findPairs(int[] nums, int k) { if (nums == null || nums.length == 0 || k < 0) { return 0; } Map<Integer, Integer> map = new HashMap<>(); int count = 0; for (int i : nums) { map.put(i, map.getOrDefault(i, 0) + 1); } for (Map.Entry<Integer, Integer> entry : map.entrySet()) { if (k == 0) { if (entry.getValue() >= 2) { count++; } } else { if (map.containsKey(entry.getKey() + k)) { count++; } } } return count; } }