Count SET Bits in a number | BIT Manipulation | C++

 How to get /count the number of ones in binary form of a number using BIT Manipulation?

 So in this blog, we will see how we can count the number of ones in a number in its binary form.
As we know that if we take AND of a number with 1 it will give 0 if the number is 0 and 1 if the number is one.
So we will take the AND every time with the LSB of the number and then right-shift the number.

Here is the code of the program in C++. The code starts in the 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;
    int ans = 0;
    while (n)
        ans += (n & 1);
        n = n >> 1;
        //OR What you can do is this
        // n = n & (n-1);
        // ans++;

int32_t main()

    clock_t z = clock();

    int t = 1;
    cin >> t;
    while (t--)

    return 0;

Thank you for reading this.

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