Problem
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.lengthi != jnums[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 <= 10^4-10^7 <= nums[i] <= 10^70 <= k <= 10^7
Solution (Java)
class Solution {
public int findPairs(int[] nums, int k) {
int res = 0;
HashSet<Integer> set = new HashSet<>();
HashSet<Integer> twice = new HashSet<>();
for (int n : nums) {
if (set.contains(n)) {
if (k == 0 && !twice.contains(n)) {
res++;
twice.add(n);
} else {
continue;
}
} else {
if (set.contains(n - k)) {
res++;
}
if (set.contains(n + k)) {
res++;
}
}
set.add(n);
}
return res;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).