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Easy

Reverse Bits

Reverse bits of a given 32 bits unsigned integer. Note: - Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned. - In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.

Examples

Input:n = 00000010100101000001111010011100
Output:964176192
The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
Input:n = 11111111111111111111111111111101
Output:3221225471
The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

Constraints

  • The input must be a binary string of length 32.

Bit by Bit Reverse

Approach

We can iterate 32 times to process all 32 bits. In each iteration `i` (from 0 to 31): 1. We extract the least significant bit of `n` using `bit = (n >> i) & 1`. 2. We shift this bit to its reversed position `31 - i` using `bit << (31 - i)`. 3. We accumulate this shifted bit into our result variable `res` using bitwise OR `|`. This effectively takes the `i`-th bit from the right of `n` and puts it at the `i`-th bit from the left of `res`.

Complexity Analysis

Time Complexity
O(1)
Space Complexity
O(1)

Time complexity is O(1) since we always do a constant number of operations (32 iterations). Space complexity is O(1).

Solution.java
public class Solution {    // you need treat n as an unsigned value    public int reverseBits(int n) {        int res = 0;                for (int i = 0; i < 32; i++) {            // Get the i-th bit of n            int bit = (n >> i) & 1;                        // Shift the bit to its reversed position and add it to res            res = res | (bit << (31 - i));        }                return res;    }}