# 1680. Concatenation of Consecutive Binary Numbers

## Description

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 109 + 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 109 + 7, the result is 505379714.


Constraints:

• 1 <= n <= 105

## Solutions

Solution 1: Bit Manipulation

By observing the pattern of number concatenation, we can find that when concatenating to the $i$-th number, the result $ans$ formed by concatenating the previous $i-1$ numbers is actually shifted to the left by a certain number of bits, and then $i$ is added. The number of bits shifted, $shift$, is the number of binary digits in $i$. Since $i$ is continuously incremented by $1$, the number of bits shifted either remains the same as the last shift or increases by one. When $i$ is a power of $2$, that is, when there is only one bit in the binary number of $i$ that is $1$, the number of bits shifted increases by $1$ compared to the last time.

The time complexity is $O(n)$, where $n$ is the given integer. The space complexity is $O(1)$.

• class Solution {
public int concatenatedBinary(int n) {
final int mod = (int) 1e9 + 7;
long ans = 0;
for (int i = 1; i <= n; ++i) {
ans = (ans << (32 - Integer.numberOfLeadingZeros(i)) | i) % mod;
}
return (int) ans;
}
}

• class Solution {
public:
int concatenatedBinary(int n) {
const int mod = 1e9 + 7;
long ans = 0;
for (int i = 1; i <= n; ++i) {
ans = (ans << (32 - __builtin_clz(i)) | i) % mod;
}
return ans;
}
};

• class Solution:
def concatenatedBinary(self, n: int) -> int:
mod = 10**9 + 7
ans = 0
for i in range(1, n + 1):
ans = (ans << i.bit_length() | i) % mod
return ans


• func concatenatedBinary(n int) (ans int) {
const mod = 1e9 + 7
for i := 1; i <= n; i++ {
ans = (ans<<bits.Len(uint(i)) | i) % mod
}
return
}

• function concatenatedBinary(n: number): number {
const mod = BigInt(10 ** 9 + 7);
let ans = 0n;
let shift = 0n;
for (let i = 1n; i <= n; ++i) {
if ((i & (i - 1n)) == 0n) {
++shift;
}
ans = ((ans << shift) | i) % mod;
}
return Number(ans);
}