To count the number of set bits of all numbers upto the number $n$ (inclusive), we can run the Brian Kernighan's algorithm on all numbers upto $n$. But this will result in a "Time Limit Exceeded" in contest submissions.

We can use the fact that for numbers upto $2^x$ (i.e. from $1$ to $2^x - 1$) there are $x \cdot 2^{x-1}$ set bits. This can be visualised as follows.

We can see that the all the columns except the leftmost have $4$ (i.e. $2^2$) set bits each, i.e. upto the number $2^3 - 1$, the number of set bits is $3 \cdot 2^{3-1}$.

With the new knowledge in hand we can come up with the following algorithm:

- Find the highest power of $2$ that is lesser than or equal to the given number. Let this number be $x$.

- Calculate the number of set bits from $1$ to $2^x - 1$ by using the formua $x \cdot 2^{x-1}$.

- Count the no. of set bits in the most significant bit from $2^x$ to $n$ and add it.

- Subtract $2^x$ from $n$ and repeat the above steps using the new $n$.

int countSetBits(int n) {

int count = 0;

while (n > 0) {

int x = std::bit_width(n) - 1;

count += x << (x - 1);

n -= 1 << x;

count += n + 1;

}

return count;

}