Count numbers with same popcount up to all-ones bound

## Problem For any positive integer `n`, define `f(n)` as the **smallest integer `>= n` whose binary representation consists only of `1` bits**. Equivalently, if `k` is the bit-length of `n` (so `2^(k-1) <= n <= 2^k - 1`), then: - `f(n) = 2^k - 1` (binary: `k` ones) You are given a positive integer `n`. Let `popcount(x)` be the number of set bits in `x`. ### Task Count how many integers `m` satisfy all of the following: - `1 <= m <= f(n)` - `m != n` - `popcount(m) = popcount(n)` Return this count. ### Input - A positive integer `n`. ### Output - An integer: the count of valid `m`. ### Constraints - `1 <= n <= 10^18` - Target time complexity: `O(log n)`. ### Examples 1) `n = 5` (binary `101`) - `k=3`, `f(n)=7` (binary `111`) - Numbers `<=7` with two set bits: `3(011), 5(101), 6(110)` - Excluding `n=5` → answer `2` 2) `n = 7` (binary `111`) - `k=3`, `f(n)=7` - Only number with three set bits is `7` itself → answer `0`