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 $O(1)$.
1 2 3 4 5 6 7 |
|
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 |
|
1 2 3 4 5 6 7 8 9 10 11 12 |
|
Solution 2
1 2 3 4 5 6 7 8 9 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
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