Sets bits from 1 to N

Problem Statement

You are gives Q queries each query containing two integers $a$ and $b$. Your task is to count the no of set-bits in for all numbers between $a$ and $b$ (both inclusive).

Example

Input: a = 1, b = 1
Output:  1

Input: a = 10, b = 15
Output: 17

Solution

Expected Time Complexity: $Log(n)$ How much the $ith$ bit in every number from $1$ to $N$ will contribute to the final answer?

It is equal to the number of numbers, in binary representation, having $ith$ bit set. which is equal to: (N / 2^i ) * 2$$^i$$^-$$^1 + X

$X = (N$ & (2^i - 1)) - 2^i$$^-$$^1+ 1

iff $X >= 0$

Time Complexity: $Log(n)$

C++

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

int AllBitsOneToN(int N) {
    int two = 2, ans = 0;
    int n = N;
    while (n) {
        ans += (N / two) * (two >> 1);
        int t = (N & (two - 1)) - (two >> 1) + 1;
        ans += max(t, 0);
        two <<= 1;
        n >>= 1;
    }
    return ans;
}

int main() {
    int q, a, b, res;
    cin >> q;
    while (q--) {
        cin >> a >> b;
        res = AllBitsOneToN(b) - AllBitsOneToN(a-1);
        cout << res << "\n";
    }
}

Python

def AllbitsOneToN(N):
    two = 2
    ans = 0
    n = N
    while n:
        ans += (N//two)*(two >> 1)
        t = (N % two) - (two >> 1) + 1
        ans += max(0, t)
        two <<= 1
        n >>= 1
    return ans

for q in range(int(input())):
    a, b = map(int, input().split())

    res = AllbitsOneToN(b)
    res -= AllbitsOneToN(a-1)

    print(res)