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Find the maximum between N and the number formed by reversing 32-bit binary representation of N

Given a positive 32-bit integer N, the task is to find the maximum between the value of N and the number obtained by decimal representation of reversal of binary representation of N in a 32-bit integer.

Examples:

Input: N = 6
Output: 1610612736
Explanation:  
Binary representation of 6 in a 32-bit integer is 00000000000000000000000000000110 i.e., (00000000000000000000000000000110)2 = (6)10.
Reversing this binary string gives (01100000000000000000000000000000)2 = (1610612736)10.
The maximum between N and the obtained number is 1610612736. Therefore, print 1610612736.

Input: N = 1610612736
Output: 1610612736

Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:  

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that obtains the number
// using said operations from N
int reverseBin(int N)
{
    // Stores the binary representation
    // of the number N
    string S = "";
    int i;
 
    // Find the binary representation
    // of the number N
    for (i = 0; i < 32; i++) {
 
        // Check for the set bits
        if (N & (1LL << i))
            S += '1';
        else
            S += '0';
    }
 
    // Reverse the string S
    reverse(S.begin(), S.end());
 
    // Stores the obtained number
    int M = 0;
 
    // Calculating the decimal value
    for (i = 0; i < 32; i++) {
 
        // Check for set bits
        if (S[i] == '1')
            M += (1LL << i);
    }
    return M;
}
 
// Function to find the maximum value
// between N and the obtained number
int maximumOfTwo(int N)
{
    int M = reverseBin(N);
 
    return max(N, M);
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << maximumOfTwo(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function that obtains the number
    // using said operations from N
    static int reverseBin(int N)
    {
        // Stores the binary representation
        // of the number N
        String S = "";
        int i;
 
        // Find the binary representation
        // of the number N
        for (i = 0; i < 32; i++) {
 
            // Check for the set bits
            if ((N & (1L << i)) != 0)
                S += '1';
            else
                S += '0';
        }
 
        // Reverse the string S
        S = (new StringBuilder(S)).reverse().toString();
 
        // Stores the obtained number
        int M = 0;
 
        // Calculating the decimal value
        for (i = 0; i < 32; i++) {
 
            // Check for set bits
            if (S.charAt(i) == '1')
                M += (1L << i);
        }
        return M;
    }
 
    // Function to find the maximum value
    // between N and the obtained number
    static int maximumOfTwo(int N)
    {
        int M = reverseBin(N);
 
        return Math.max(N, M);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int N = 6;
        System.out.print(maximumOfTwo(N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function that obtains the number
# using said operations from N
def reverseBin(N):
     
    # Stores the binary representation
    # of the number N
    S = ""
    i = 0
 
    # Find the binary representation
    # of the number N
    for i in range(32):
         
        # Check for the set bits
        if (N & (1 << i)):
            S += '1'
        else:
            S += '0'
 
    # Reverse the string S
    S = list(S)
    S = S[::-1]
    S = ''.join(S)
     
    # Stores the obtained number
    M = 0
 
    # Calculating the decimal value
    for i in range(32):
         
        # Check for set bits
        if (S[i] == '1'):
            M += (1 << i)
             
    return M
 
# Function to find the maximum value
# between N and the obtained number
def maximumOfTwo(N):
     
    M = reverseBin(N)
 
    return max(N, M)
 
# Driver Code
if __name__ == '__main__':
     
    N = 6
    print(maximumOfTwo(N))
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
 
class GFG{
static string ReverseString(string s)
    {
        char[] array = s.ToCharArray();
        Array.Reverse(array);
        return new string(array);
    }
   
// Function that obtains the number
// using said operations from N
public static int reverseBin(int N)
{
   
    // Stores the binary representation
    // of the number N
    string S = "";
    int i;
 
    // Find the binary representation
    // of the number N
    for (i = 0; i < 32; i++) {
 
        // Check for the set bits
        if ((N & (1L << i)) != 0)
            S += '1';
        else
            S += '0';
    }
 
    // Reverse the string S
    S = ReverseString(S);
     
 
    // Stores the obtained number
    int M = 0;
 
    // Calculating the decimal value
    for (i = 0; i < 32; i++) {
 
        // Check for set bits
        if (S[i] == '1')
            M +=  (1 << i);
    }
    return M;
}
 
// Function to find the maximum value
// between N and the obtained number
static int maximumOfTwo(int N)
{
    int M = reverseBin(N);
 
    return Math.Max(N, M);
}
 
// Driver Code
static void Main()
{
    int N = 6;
    Console.Write(maximumOfTwo(N));
 
}
}
 
// This code is contributed by SoumikMondal

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function that obtains the number
// using said operations from N
function reverseBin(N)
{
    // Stores the binary representation
    // of the number N
    let S = "";
    let i;
 
    // Find the binary representation
    // of the number N
    for (i = 0; i < 32; i++) {
 
        // Check for the set bits
        if (N & (1 << i))
            S += '1';
        else
            S += '0';
    }
 
    // Reverse the string S
    S = S.split("").reverse().join("");
 
    // Stores the obtained number
    let M = 0;
 
    // Calculating the decimal value
    for (i = 0; i < 32; i++) {
 
        // Check for set bits
        if (S[i] == '1')
            M += (1 << i);
    }
    return M;
}
 
// Function to find the maximum value
// between N and the obtained number
function maximumOfTwo(N)
{
    let M = reverseBin(N);
 
    return Math.max(N, M);
}
 
// Driver Code
    let N = 6;
    document.write(maximumOfTwo(N));
 
</script>

Output: 

1610612736

 

Time Complexity: O(32)
Auxiliary Space: O(1)

 


 


Reffered: https://www.geeksforgeeks.org

Bit Magic

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Type:
 Geek
Category:
 Coding
Sub Category:
 Tutorial
Uploaded by:
 Admin
Views:
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