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K-diff Pairs in an Array in C++
Suppose we have an array and an integer k, we have to find the number of unique k-diff pairs in the array. Here the k-diff pair is like (i, j), where i and j are both are present in the array and their absolute difference is k.
So, if the input is like [3,1,4,1,5], k = 2, then the output will be 2, as there are two 2-diff pairs in the array-like (1,3) and (3,5).
To solve this, we will follow these steps −
Define maps called seen and done
Define one set s
if k < 0, then −
return 0
for initialize i := 0, when i < size of nums, update (increase i by 1), do −
(increase seen[nums[i]] by 1)
insert nums[i] into s
ans := 0
for each element it in s, do −
if k is same as 0, then −
if seen[it] > 1, then −
(increase ans by 1)
Otherwise
increase done[it] by 1
if (it + k) is in seen but not in done, then −
(increase ans by 1)
if (it - k) is in seen but not in done, then −
(increase ans by 1)
return ans
Example
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h&g;
using namespace std;
class Solution {
public:
int findPairs(vector<int>& nums, int k) {
map<int, int> seen, done;
set<int> s;
if (k < 0)
return 0;
for (int i = 0; i < nums.size(); i++) {
seen[nums[i]]++;
s.insert(nums[i]);
}
int ans = 0;
for (auto it = s.begin(); it != s.end(); it++) {
if (k == 0) {
if (seen[*it] > 1)
ans++;
}
else {
done[*it]++;
if (seen.find(*it + k) != seen.end() && done.find(*it + k) == done.end())
ans++;
if (seen.find(*it - k) != seen.end() && done.find(*it - k) == done.end())
ans++;
}
}
return ans;
}
};
main(){
Solution ob;
vector<int> v = {3,1,4,1,5};
cout << (ob.findPairs(v, 2));
}Input
{3,1,4,1,5}, 2Output
2
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