Leetcode Problem-190 [Type -> easy]:

Description:

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.

Implementation in java:

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        return Integer.reverse(n);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNextInt()) {
            int n = sc.nextInt();
            Solution example = new Solution();
            int result = example.reverseBits(n);
            System.out.println(result);
        }
        sc.close();
    }
}

Approach:

• 1. Imports and Class Declaration:

  import java.util.Scanner;
  public class Solution {

  Imports the Scanner class for input.

  Declares the Solution class.

• 2. reverseBits Method:

  public int reverseBits(int n) {
    return Integer.reverse(n);
  }

  This method uses the Integer.reverse(n) function, which reverses the bits of the given 32-bit integer n.

• 3. main Method:

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    while (sc.hasNextInt()) {
        int n = sc.nextInt();
        Examples example = new Examples();
        int result = example.reverseBits(n);
        System.out.println(result);
    }
    sc.close();
  }

  Scanner Initialization: Creates a Scanner object sc to read input from the user.

  Loop for Input: Uses while (sc.hasNextInt()) to continually read integers as long as there are any.

  Instance Creation and Method Call: For each integer input, it creates an instance of the Solution class and calls the reverseBits method on it.

  Print Result: Outputs the reversed bits result.

  Close Scanner: Ensures the Scanner is closed after usage.

Time and Space Complexity

Time Complexity

• $O(1)$ - The Integer.reverse() method in Java performs a fixed number of bit operations (at most 32 operations for a 32-bit integer). Since the input size is fixed, the algorithm runs in constant time.

Space Complexity

• $O(1)$ - The algorithm uses only a constant amount of extra space, regardless of input size. It operates directly on the bit representation without requiring additional data structures.

Explanation

Reversing bits of a 32-bit integer is fundamentally a fixed-size operation. Java's Integer.reverse() efficiently reverses the bit sequence by performing bitwise operations. The algorithm swaps bits from both ends moving toward the center. Since we're dealing with a fixed 32-bit representation, both time and space complexity are constant.

Submission details: