1class Solution {
2    public int countMonobit(int n) {
3        // ans counts how many "all-ones" binary numbers (1, 3, 7, 15, ...) fit within n,
4        // starting at 1 to account for the base case.
5        int ans = 1;
6
7        // threshold represents numbers of the form 2^k - 1: 1, 3, 7, 15, ...
8        // i drives the bit shift; each iteration adds the next power of two.
9        for (int i = 1, threshold = 1; threshold <= n; ++i) {
10            // Count this all-ones number since it does not exceed n.
11            ++ans;
12
13            // Advance threshold from (2^i - 1) to (2^(i+1) - 1) by adding 2^i.
14            threshold += (1 << i);
15        }
16
17        return ans;
18    }
19}
20```
21
22**A few notes on alternative perspectives:**
23
241. **Naming choice:** I renamed `x` to `threshold` since it represents the upper bound being tested. You could also name it `allOnes` to emphasize its binary pattern (`1, 11, 111, ...`).
25
262. **Potential overflow caution:** When `i` reaches 31, `1 << i` can overflow into negative values, and `threshold` could overflow too. For large `n`, a safer approach uses the bit length directly:
27
28```java
29class Solution {
30    public int countMonobit(int n) {
31        // The number of all-ones values (1, 3, 7, ...) that are <= n
32        // equals the bit length of n; add 1 to match the original base case.
33        return 32 - Integer.numberOfLeadingZeros(n) + 1;
34    }
35}
36

1class Solution {
2public:
3    int countMonobit(int n) {
4        // Count starts at 1 to account for the base case (value 1, i.e., 2^0)
5        int count = 1;
6
7        // 'value' represents numbers of the form (2^k - 1): 1, 3, 7, 15, ...
8        // which are binary numbers consisting entirely of consecutive 1-bits.
9        // 'bitIndex' tracks the next power of two to add into 'value'.
10        for (int bitIndex = 1, value = 1; value <= n; ++bitIndex) {
11            // Each qualifying value increments the count.
12            ++count;
13
14            // Add the next power of two (2^bitIndex) to extend the all-ones pattern.
15            value += (1 << bitIndex);
16        }
17
18        return count;
19    }
20};
21```
22
23**A few observations and alternative perspectives:**
24
251. **Potential overflow concern:** When `n` is large (close to `INT_MAX`), the expression `value += (1 << bitIndex)` can overflow a signed 32-bit `int`, causing undefined behavior. A safer variant uses `long long` for `value` and the shift:
26
27```cpp
28class Solution {
29public:
30    int countMonobit(int n) {
31        int count = 1;
32        // Use long long to avoid signed overflow when n is near INT_MAX.
33        for (int bitIndex = 1; (1LL << (bitIndex + 1)) - 1 <= n; ++bitIndex) {
34            ++count;
35        }
36        return count;
37    }
38};
39```
40
412. **Mathematical perspective:** Since the qualifying values are exactly `2^k - 1`, the answer equals the number of bits in `n` (i.e., the count of all-ones numbers not exceeding `n`). This can be computed directly:
42
43```cpp
44class Solution {
45public:
46    int countMonobit(int n) {
47        // The largest all-ones number <= n has the same bit-length as n.
48        // Equivalent to counting how many of 1, 3, 7, 15, ... are <= n.
49        int count = 0;
50        while (n > 0) {
51            ++count;
52            n >>= 1; // Drop one bit each iteration.
53        }
54        return count;
55    }
56};
57

1/**
2 * Counts a value based on consecutive "all-ones" binary numbers (1, 3, 7, 15, ...)
3 * that do not exceed the given limit n.
4 *
5 * @param n - The upper bound to compare against the accumulated value.
6 * @returns The computed count.
7 */
8function countMonobit(n: number): number {
9    // Start the answer at 1 to account for the base case.
10    let answer = 1;
11
12    // `index` tracks the current bit position (1, 2, 3, ...).
13    // `value` accumulates all-ones binary numbers: 1, 3, 7, 15, ...
14    for (let index = 1, value = 1; value <= n; ++index) {
15        // For every all-ones value within the limit, increment the answer.
16        ++answer;
17
18        // Add the next power of two (2^index) to extend `value`
19        // to the next all-ones binary number.
20        value += 1 << index;
21    }
22
23    return answer;
24}
25