Reverse Bits

The input n = 43261596 is a 32-bit unsigned integer, so write it with all 32 positions, leading zeros included: $00000010100101000001111010011100$ . Counting position $0$ as the least significant bit, its 1s sit at positions $2, 3, 4, 7, 9, 10, 11, 12, 18, 20, 23, 25$ .
Reversing the bits sends the bit at position $i$ to position $31 - i$ : position $2 \to 29$ , $3 \to 28$ , $25 \to 6$ , and so on. The reversed pattern is $00111001011110000010100101000000$ .
Reading that back as a number: $2^{29} + 2^{28} + 2^{27} + 2^{24} + 2^{22} + 2^{21} + 2^{20} + 2^{19} + 2^{13} + 2^{11} + 2^{8} + 2^{6} = 964176192$ .
Notice the leading zeros matter: the input's 6 leading zeros become the output's 6 trailing zeros. Reversing only the 26 significant digits would give a much smaller, wrong number.
Another check: $4294967293$ is all 1s except position $1$ , so its single 0 moves to position $30$ , giving $2^{32} - 1 - 2^{30} = 3221225471$ ; and $0$ stays $0$ .

Background Knowledge

Bit Manipulation is a fundamental concept in computer science that involves performing operations on the binary representation of numbers. In this problem, we're dealing with a 32-bit unsigned integer, which means it's represented by 32 binary digits (bits). To reverse the bits of a number, we need to understand how to manipulate these bits using bitwise operators. The key operators used in bit manipulation are: & (bitwise AND), | (bitwise OR), ^ (bitwise XOR), ~ (bitwise NOT), << (left shift), and >> (right shift).

The binary representation of a number is crucial in bit manipulation. For example, the decimal number 10 is represented as 1010 in binary. To reverse the bits of a number, we need to understand how to extract and manipulate individual bits. This can be done using bitwise operators and shift operators. For instance, the expression x & 1 extracts the least significant bit (LSB) of x, while x >> 1 shifts the bits of x one position to the right.

In the context of this problem, we need to reverse the bits of a 32-bit unsigned integer. This means we need to take the binary representation of the number, reverse the order of its bits, and return the resulting integer. For example, if the input is 00000000000000000000000000001010, the output should be 01010000000000000000000000000000. To achieve this, we'll need to use a combination of bitwise operators and shift operators to manipulate the bits of the input number.

Algorithm/Approach

The general approach to solving this problem involves using a loop to iterate over the bits of the input number. In each iteration, we'll extract the least significant bit (LSB) of the input number and append it to the result. We'll then shift the bits of the input number one position to the right to move to the next bit. This process will continue until all 32 bits have been processed.

Step-by-Step Strategy

To implement the solution, follow these steps:

Initialize a variable to store the result.
Loop 32 times to process each bit of the input number.
In each iteration, extract the LSB of the input number using a bitwise operator.
Append the extracted bit to the result using a shift operator.
Shift the bits of the input number one position to the right to move to the next bit.
After the loop, return the result.

Common Pitfalls

When implementing the solution, watch out for the following common pitfalls:

Forgetting to initialize the result variable to 0.
Using the wrong bitwise operator to extract the LSB of the input number.
Shifting the bits of the result in the wrong direction.
Not looping the correct number of times (32 times for a 32-bit integer).

Time & Space Complexity

The expected time complexity of the solution is O(32), which simplifies to O(1) since the number of iterations is constant. The space complexity is O(1) since we only use a constant amount of space to store the result and other variables.

Example

Constraints

More from Blind 75

Background Knowledge

Algorithm/Approach

Step-by-Step Strategy

Common Pitfalls

Time & Space Complexity