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.


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Example 1:

Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.


Example 2:

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.


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Constraints:


	The input must be a binary string of length 32


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Follow up: If this function is called many times, how would you optimize it?

class Solution:
# @param n, an integer
# @return an integer
def reverseBits(self, n):
    res=0
    for i in range(32):
        bit=(n>>i)&1
        res=res|bit<<(31-i)
    return res
public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int mask = 0;
        int smask = 0;
        int j = 0;
        int rev = 0;
        
        // basically we are checking that the number is set bit or not 
        // if the number is set bit then we are appending that to our main answer i.e, rev
        for(int i=31 ; i>=0 ; i--){
            mask = 1<<i;
            if((mask&n)!=0){
                smask = 1<<j;
                rev = rev|smask;
            }
            j++;
        }
        
        // Time Complexity : O(32 for int)
        // Space Complexity : O(1)
        return rev;
    }
}
class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        int ans =0;
        int i =1;
        int bit =0;
        while(i<32){
            bit = n&1;
            ans = ans|bit;
            n = n>>1;
            ans = ans<<1;
            i++;
        }
        if (i ==32){
            bit = n&1;
            ans = ans|bit;
        }
        return ans;
    }
};
/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */

// remember the binary must always be of length 32 ;);
var reverseBits = function(n) {
    const reversedBin = n.toString(2).split('').reverse().join('');
    const result =  reversedBin.padEnd(32,'0'); 
    return parseInt(result, 2);
};
Reverse Bits