# 190. Reverse Bits

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Reverse bits of a given 32 bits unsigned integer.

**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**.

Approach 1:

*Bit manipulation*

The approach is simple .We take the least significant bit in ‘n’ followed by right shifting it. The obtained bit is added at the corresponding position in the reversed form(This is done by adding the obtained bit at the LSB of the intermediated result followed by a left shift). At the end of the iteration we get the required answer in the reverse form

`class Solution:`

def reverseBits(self, n: int) -> int:

res=0

for i in range(0,32):

res<<=1

res|=(n&1)

n>>=1

return res

Time :O(1)

Space : O(1)