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In this post, you will find the solution for the **Reverse Bits** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Reverse Bits– LeetCode Problem

Reverse Bits– LeetCode Problem

**Problem:**

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`

.

**Example 1:**

Input:n = 00000010100101000001111010011100Output:964176192 (00111001011110000010100101000000)Explanation:The input binary string00000010100101000001111010011100represents the unsigned integer 43261596, so return 964176192 which its binary representation is00111001011110000010100101000000.

**Example 2:**

Input:n = 11111111111111111111111111111101Output:3221225471 (10111111111111111111111111111111)Explanation:The input binary string11111111111111111111111111111101represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is10111111111111111111111111111111.

**Constraints:**

- The input must be a
**binary string**of length`32`

Reverse Bits– LeetCode Solutions

Reverse BitsSolution in C++:

class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t ans = 0; for (int i = 0; i < 32; ++i) if (n >> i & 1) ans |= 1 << 31 - i; return ans; } };

Reverse BitsSolution in Java:

public class Solution { // you need treat n as an unsigned value public int reverseBits(int n) { int ans = 0; for (int i = 0; i < 32; ++i) if ((n >> i & 1) == 1) ans |= 1 << 31 - i; return ans; } }

Reverse BitsSolution in Python:

class Solution: def reverseBits(self, n: int) -> int: ans = 0 for i in range(32): if n >> i & 1: ans |= 1 << 31 - i return ans