In this HackerEarth Monk and Tasks problem solution, Monk A loves to complete all his tasks just before the deadlines for introducing unwanted thrill in his life. But, there is another Monk D who hates this habit of Monk A and thinks it's risky.

To test Monk A, Monk D provided him tasks for N days in the form of an array Array, where the elements of the array represent the number of tasks.

The number of tasks performed by Monk A on an ith day is the number of ones in the binary representation of Array.

Monk A is fed up of Monk D, so to irritate him even more, he decides to print the tasks provided in non-decreasing order of the tasks performed by him each day. Help him out!

## HackerEarth Monk and Tasks problem solution.

```#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX = 1000005;
ll Ar[MAX];
vector <ll> vec;
//counting number of ones in binary representation.
ll count_one(ll a)
{
ll coun=0;
while(a>0)
{
coun++;
a = a&(a-1);
}
return coun;
}
int main()
{

int t;
for(cin>>t;t;--t)
{
ll n;
cin>>n;
ll maxi=-1;
for(int i=0;i<=70;i++)
vec[i].clear();
for(int i=0;i<n;i++)
{
cin>>Ar[i];
ll num_one = count_one(Ar[i]);
vec[num_one].push_back(Ar[i]);
if(num_one > maxi)
maxi=num_one;
}
for(int i=0;i<=maxi;i++)
{
for(int j=0;j<vec[i].size();j++)
{
if(i == maxi and j==vec[i].size()-1)
{
cout<<vec[i][j]<<endl;
}
else
{
cout<<vec[i][j]<<" ";
}
}
}

}
return 0;
}```

### Second solution

```#include <bits/stdc++.h>
using namespace std;

const int MAX = 1e5 + 5;
pair <int, pair<int, long long> > p[MAX];

int main() {
int t, n, x;
long long a, b;
cin >> t;
assert(1<= t and t <= 100);
while(t--) {
cin >> n;
assert(1 <= n and n <= 100000);
for(int i = 0; i<n; ++i) {
cin >> a;
assert(1 <= a and a <= 1000000000000000000LL);
x = 0;
b = a;
while(b)
x += (b & 1), b >>= 1;
p[i] = make_pair(x, make_pair(i, a));
}
sort(p, p + n);
for(int i = 0;i < n;++i)
cout << p[i].second.second << ' ';
cout << endl;
}
return 0;
}```