13(April) Reverse Bits

13. Reverse Bits

The problem can be found at the following link: Question Link

Problem Description

Given a number ( x ), reverse its binary form and return the answer in decimal.

Example 1:

Input:

x = 1

Output:

2147483648

Explanation: Binary of 1 in 32 bits representation-

00000000000000000000000000000001

Reversing the binary form we get,

10000000000000000000000000000000

whose decimal value is 2147483648.

My Approach

To reverse the bits of a number, we can iterate through each bit of the number from right to left (least significant bit to most significant bit) and build the reversed number bit by bit.

1. Initialization:

• Initialize a variable ans to store the reversed bits. Initially, set it to 0.

1. Bit Reversal:

• Iterate from i = 0 to 31 (assuming a 32-bit integer).

• In each iteration:

  • Shift the ans left by 1 (multiply by 2) to make space for the next bit.

  • Take the least significant bit of x using the bitwise AND operation (x & 1), and OR it with ans.

  • Shift x right by 1 to process the next bit.

1. Return:

• Return the ans as the reversed binary form of x in decimal.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(log (x)), as the number of iterations depends on the number of bits in (x).

• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space.

Code (C++)

class Solution {
public:
    long long reversedBits(long long x) {
        long long ans = 0;
        for (int i = 0; i < 32; ++i) {
            ans = (ans << 1) | (x & 1); // Shift ans left by 1 and OR it with the least significant bit of x
            x >>= 1; // Shift x right by 1 to process the next bit
        }
        return ans;
    }
};

Contribution and Support

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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