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
public class Solution {
public int FindPairs(int[] nums, int k) {
if(nums.Length ==0){
return 0;
}
var map = new Dictionary<int,int>();
for(int i=0;i<nums.Length;i++){
int val=0;
map.TryGetValue(nums[i],out val);
map[nums[i]]= val+1;
}
int count =0;
foreach(var item in map){
if(k==0){
if(item.Value>1){
count++;
}
}
else{
if(map.ContainsKey(item.Key-k)){
count++;
}
}
}
return count;
}
}
Time Complexity: O(n)
Space Complexity: O(n)