How to check if a number is the power of two using BIT Manipulation?
We will see here how we can check whether a number is the power of two using BIT Manipulation.
So what we can do is we can convert the given number into Binary.
Any number which is the power of 2 in binary will look like 10000...000 with only one 1 at the beginning and then only zero.
Now we will convert the given number - 1 into binary and it will look like 01111...111 means it will become one's complement of the given number.
Now if we take and of both the number we will get 0 and hence we can know that the given number is a power of 2.
Here is the code in C++.
The code starts from Solve function.
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define F first
#define S second
#define pb push_back
#define si set <int>
#define vi vector <int>
#define pii pair <int, int>
#define vpi vector <pii>
#define vpp vector <pair<int, pii>>
#define mii map <int, int>
#define mpi map <pii, int>
#define spi set <pii>
#define endl "\n"
#define sz(x) ((int) x.size())
#define all(p) p.begin(), p.end()
#define double long double
#define que_max priority_queue <int>
#define que_min priority_queue <int, vi, greater<int>>
#define bug(...) __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a) for(auto x : a) cout << x << " "; cout << endl
#define print1(a) for(auto x : a) cout << x.F << " " << x.S << endl
#define print2(a,x,y) for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl
void solve() {
int n;
cin>>n;
if((n & (n-1))==0){
cout<<"The number is power of 2"<<endl;
}
else{
cout<<"The number is Not power of 2"<<endl;
}
}
int32_t main()
{
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
clock_t z = clock();
int t = 1;
// cin >> t;
while (t--) solve();
return 0;
}
Thank you for reading this.
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