Largest number M having bit count of N such that difference between their OR and XOR value is maximized

Given a natural number N, the task is to find the largest number M having the same length in binary representation as N such that the difference between N | M and N ^ M is maximum.

Examples:

Input: N = 6
Output: 7
Explanation:  
All number numbers having same length in binary representation as N are 4, 5, 6, 7.
(6 | 4) – (6 ^ 4) = 4
(6 | 5) – (6 ^ 5) = 4
(6 | 6) – (6 ^ 6) = 6
(6 | 7) – (6 ^ 7) = 6
Hence, largest M for which (N | M) – (N ^ M) is maximum is 7

Input: N = 10
Output: 15
Explanation:  
The largest number M = 15 which has the same length in binary representation as 10 and the difference between N | M and N ^ M is maximum.

Naive Approach: The idea is to simply find all the numbers having the same length in binary representation as N and then for every number iterate and find the largest integer having (N | i) – (N ^ i) maximum. 



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

Efficient Approach: The idea is to initialize M = 0 and iterate bit by bit in N (say i) and set or unset the ith bit of M according to the following 2 observations :

Below is the implementation of the above approach:

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
int maxORminusXOR(int N)
{
    // Find the most significant
    // bit of N
    int MSB = log2(N);
  
    // Initialize M
    int M = 0;
  
    // Set all the bits of M
    for (int i = 0; i <= MSB; i++)
        M += (1 << i);
  
    // Return the answer
    return M;
}
  
// Driver Code
int main()
{
    // Given Number N
    int N = 10;
  
    // Function Call
    cout << maxORminusXOR(N);
    return 0;
}
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// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
static int maxORminusXOR(int N)
{
      
    // Find the most significant
    // bit of N
    int MSB = (int)Math.ceil(Math.log(N));
  
    // Initialize M
    int M = 0;
  
    // Set all the bits of M
    for(int i = 0; i <= MSB; i++)
        M += (1 << i);
  
    // Return the answer
    return M;
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given number N
    int N = 10;
  
    // Function call
    System.out.print(maxORminusXOR(N));
}
}
  
// This code is contributed by Rajput-Ji
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# Python3 program for the above approach
import math
  
# Function to find the largest number
# M having the same length in binary
# form as N such that the difference
# between N | M and N ^ M is maximum
def maxORminusXOR(N):
  
    # Find the most significant
    # bit of N
    MSB = int(math.log2(N));
  
    # Initialize M
    M = 0
  
    # Set all the bits of M
    for i in range(MSB + 1):
        M += (1 << i)
  
    # Return the answer
    return M
  
# Driver code
if __name__ == '__main__'
      
    # Given Number N
    N = 10
  
    # Function call
    print(maxORminusXOR(N))
  
# This code is contributed by jana_sayantan
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// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
static int maxORminusXOR(int N)
{
      
    // Find the most significant
    // bit of N
    int MSB = (int)Math.Ceiling(Math.Log(N));
  
    // Initialize M
    int M = 0;
  
    // Set all the bits of M
    for(int i = 0; i <= MSB; i++)
        M += (1 << i);
  
    // Return the answer
    return M;
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given number N
    int N = 10;
  
    // Function call
    Console.Write(maxORminusXOR(N));
}
}
  
// This code is contributed by 29AjayKumar
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Output: 
15

Time Complexity: O(log N)
Auxiliary Space: O(1)

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