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

  • Difficulty Level : Easy
  • Last Updated : 20 Apr, 2021

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 :

  • When an ith bit of N is set: In this case, if we unset the ith bit of M, ith bit of both N | M and N^M will be set whereas on setting this bit of M, an ith bit of N|M will be set and N^M will be unset which will increase (N | M) – (N ^ M). Hence, it is optimal to set this bit of M.
  • When an ith bit of N is unset: In this case, if we set this bit of M, both N|M and N^M will have this bit set or on keeping this bit of M unset both N|M and N^M will have this bit unset. So, in this case, we cannot increase the difference between them but as the requirement is to output the maximum M possible, so set this bit of M.
  • From the above observations, it is clear that M will have all the bits set.

Below is the implementation of the above approach:

C++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
 
// JavaScript implementation of the above approach
  
// 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
function maxORminusXOR(N)
{
       
    // Find the most significant
    // bit of N
    let MSB = Math.ceil(Math.log(N));
   
    // Initialize M
    let M = 0;
   
    // Set all the bits of M
    for(let i = 0; i <= MSB; i++)
        M += (1 << i);
   
    // Return the answer
    return M;
}
 
// Driver code
         
    // Given number N
    let N = 10;
   
    // Function call
    document.write(maxORminusXOR(N));
   
  // This code is contributed by code_hunt.
</script>
Output: 
15

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

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