# 190. Reverse Bits(Solution || Leetcode easy || Java)

2 min readDec 20, 2022

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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 = 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.

**Constraints:**

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

# SOLUTION:

`public class Solution {`

// you need treat n as an unsigned value

public int reverseBits(int n) {

//Create a integer to store answer

int ans = 0;

//run the loop 32 times as it is given in constraints that

//the int is 32 bit long

for(int i = 0; i < 32; i++){

//append the last bit of integer to reverse it

ans |= (n & 1);

//this condition is necessary anotherwise the loop will

// add a extra zero

if(i != 31)

//left shift ans to store next bit

ans = ans << 1;

//as the last bit is stored in ans remove that bit

n = n >> 1;

}

return ans;

}

}

Runtime 1 ms

Beats

97.45%

Memory 42.6 MB

Beats

34.36%

Thank you for reading. If you have any queries then, please let me know in the comment section. I will surely be responsive toward it.