#include
#include
using namespace std;
int main() {
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
int key;
cin>>key;
sort(arr,arr+n);
int i=0;int j=n-1;
while(i<j)
{
if(arr[i]+arr[j]==key)
{
cout<<arr[i]<<" and "<<arr[j]<<endl;
i++;j++;
}
if(arr[i]+arr[j]>key)
{
j–;
}
else{
i++;
}
}
return 0;
}
test case 2 is note working why???
code is similar to solution.
Hello Utkarsh, just don’t do
either do i++ or j-- (so that every pair is taken care of)
But if i take the case as
10
2 2 2 4 4 4 4 4 4 4
sum=6
than total pairs should be 21 times 2 4…
but gives only 7 times of it…
Yes you have a valid point!
This cannot be dealt in O(n) complexity!
instead you should iterate for every element, and check for valid sum across all elements which brings complexity to O(n^2).
Pseudo Code would be:
for i: 0 to n-1
for j: i+1 to n-1
if(a[i]+a[j]==sum)
print a[i],a[j]
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