# 1680. Concatenation of Consecutive Binary Numbers

## Bit manipulation, mathematics and recursion

### Problem statement

Given an integer `n`

, return the decimal value of the binary string formed by concatenating the binary representations of `1`

to `n`

in order, modulo `10^9 + 7`

.

#### Example 1

```
Input: n = 1
Output: 1
Explanation: "1" in binary corresponds to the decimal value 1.
```

#### Example 2

```
Input: n = 3
Output: 27
Explanation: In binary, 1, 2, and 3 corresponds to "1", "10", and "11".
After concatenating them, we have "11011", which corresponds to the decimal value 27.
```

#### Example 3

```
Input: n = 12
Output: 505379714
Explanation: The concatenation results in "1101110010111011110001001101010111100".
The decimal value of that is 118505380540.
After modulo 10^9 + 7, the result is 505379714.
```

#### Constraints

`1 <= n <= 10^5`

.

### Solution: Recursive

There must be some relationship between the result of `n`

and the result of `n - 1`

.

First, let us list some first values of `n`

.

For

`n = 1`

: the final binary string is`"1"`

, its decimal value is`1`

.For

`n = 2`

: the final binary string is`"110"`

, its decimal value is`6`

.For

`n = 3`

: the final binary string is`"11011"`

, its decimal value is`27`

.

Look at `n = 3`

, you can see the relationship between the decimal value of `"11011"`

and the one of `"110"`

(of `n = 2`

) is:

```
27 = 6 * 2^2 + 3
Dec("11011") = Dec("110") * 2^num_bits("11") + Dec("11")
Result(3) = Result(2) * 2^num_bits(3) + 3.
```

The same equation for `n = 2`

:

```
6 = 1 * 2^2 + 2
Dec("110") = Dec("1") * 2^num_bits("10") + Dec("10")
Result(2) = Result(1) * 2^num_bits(2) + 2.
```

In general, the recursive relationship between `n`

and `n - 1`

is:

```
Result(n) = Result(n - 1) * 2^num_bits(n) + n.
```

#### Code

```
#include <cmath>
#include <iostream>
int concatenatedBinary(int n) {
unsigned long long result = 1;
for (int i = 2; i <= n; i++) {
const int num_bits = std::log2(i) + 1;
result = ((result << num_bits) + i) % 1000000007;
}
return result;
}
int main() {
std::cout << concatenatedBinary(1) << std::endl;
std::cout << concatenatedBinary(3) << std::endl;
std::cout << concatenatedBinary(12) << std::endl;
}
```

```
Output:
1
27
505379714
```

#### Complexity

Runtime:

`O(n)`

.Extra space:

`O(1)`

.

### References

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