In this HackerEarth Fredo and Sums problem solution, Fredo has an array A consisting of N elements. He wants to divide the array into N/2 pairs where each array element comes in exactly one pair. Say that pair i has elements Xi and Yi, he defines S as :

S = Sigma(i=1,N/2) abs(Xi - Yi)

He asks you to find the minimum and maximum value of S.


HackerEarth Fredo and Sums problem solution


HackerEarth Fredo and Sums problem solution.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
int t;
cin>>t;
assert(t>=1 && t<=10);
while(t--)
{
int n,a[100005];
cin>>n;
assert(n>=1 & n<=1e5);
assert(!(n&1));
for(int i=0;i<n;i++)
{
cin>>a[i];
assert(a[i]>=-1e9 && a[i]<=1e9);
}
sort(a,a+n);
ll mins=0,maxs=0;
for(int i=1;i<n;i+=2)
mins+=(ll)(a[i]-a[i-1]);
for(int i=0;i<n/2;i++)
maxs+=(ll)(a[n-1-i]-a[i]);
cout<<mins<<" "<<maxs<<"\n";
}
return 0;
}