I think we can using digit dp on binary representation of number

You Can Take Help From this

They Explain all way to explain this problem.

Digit Dp and maybe Gray code can be used. https://www.geeksforgeeks.org/what-is-gray-code/

Use the fact that the bits at ith position alternate after every 2^i numbers. So for a bit i, find the first number after L which has it set call it L_set, and first number below L which has it un-set called it L_unset this can be done in constant time. Similarly do it for R also, now that you know the range — [L_unset+1, R_unset-1] has a multiple of 2^i numbers and you know L_unset, L_set, R_unset and R_set you can subtract the out of range elements from this count.

For i. bit,

$$$f(X)$$$ will give count of numbers in $$$1 \leq y \leq X$$$

You can see the contrubition pattern is $$$i$$$ number of 0s, $$$i$$$ number of 1s, $$$i$$$ number of 0s, etc.

So if $$$X$$$ equals $$$2^{(i + 1)} \times y$$$, count is $$$2^i \times y$$$

Else, divide it into 2 parts: $$$2^{(i + 1)} \times y$$$, and another part, $$$Z$$$, which is in $$$0 \leq Z \le 2^{(i + 1)}$$$, and the first part calculation is the same.

We can say for another part their contrubition pattern is $$$i$$$ numbers of 0 and $$$i$$$ number of 1. (Not any more)

If it is smaller than $$$2^i$$$, its contribution is 0

Else its contrubition is $$$Z - 2^i$$$

And answer for i. bit is $$$f(R) - f(L - 1)$$$

You can use the fact that the first bit is going to appear 1 time every 2 numbers, the second bit is going to appear 2 times every 4 number, the third bit 4 times every 8 numbers and so on

Hi, you can try this. Given function computes total set bits from 1 to A. You can either use it as f(R) — f(L-1) or modify this function itself to compute it directly from L->R.

It runs in O(32) and uses the fact that at a given bit position, there's always a repetitive pattern. I'll leave the rest to you for figuring out.

    def solve(A):
        count = 0
        
        for b in range(32):
            x = 1<<b
                
            y = A-x+1
            if y<0:
                break
                
            z = y//x
            
            count += (z//2+z%2)*x + (z%2==0)*(y%x)
                        
        return count

I recently created a video on this topic. I also created a practice problem. You can submit your code here

This stack overflow link might be helpful -> Stack overflow question