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|nums[i] - nums[j]| == kNotice 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] <= 1070 <= k <= 107class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_map<int,int> m;
for(int i:nums)
m[i]++;
int res=0;
for(auto e:m){
if(k==0){
if(e.second>1) // same element present twice, hence 0-diff holds
res++;
}
else if(m.find(e.first+k)!=m.end()) // if k-diff found
res++;
}
return res;
}
};