Skip to content
Related Articles

Related Articles

Improve Article

Find a number M < N such that difference between their XOR and AND is maximum

  • Last Updated : 28 Sep, 2021

Given a natural number N, the task is to find a number M smaller than N such that the difference between their bitwise XOR (N ^ M) and bitwise AND (N & M) is maximum.
 

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: N = 4
Output: 3
Explanation:  
(4 ^ 0) – (4 & 0) = 4
(4 ^ 1) – (4 & 1) = 5
(4 ^ 2) – (4 & 2) = 6
(4 ^ 3) – (4 & 3) = 7
Hence, the value of M is 3.



Input: N = 6
Output: 1
Explanation: 
The difference between N ^ M and N & M is maximum when M = 1.

Naive Approach: The idea is to iterate for every element less than N and find M for which N^M – N&M is maximum. Below are the steps:
 

  • Initialize a variable let’s say maxDiff with 0 and M with -1.
  • Iterate from 0 to N-1 and calculate diff = N^i -N&i.
  • If diff is greater or equal to maxDiff assign M = i and maxDiff = diff.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return M<N such that
// N^M - N&M is maximum
int getMaxDifference(int N)
{
    // Initialize variables
    int M = -1;
    int maxDiff = 0;
 
    // Iterate for all values < N
    for (int i = 0; i < N; i++) {
 
        // Find the difference between
        // Bitwise XOR and AND
        int diff = (N ^ i) - (N & i);
 
        // Check if new difference is
        // greater than previous maximum
        if (diff >= maxDiff) {
 
            // Update variables
            maxDiff = diff;
            M = i;
        }
    }
 
    // Return the answer
    return M;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 6;
 
    // Function Call
    cout << getMaxDifference(N);
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to return M<N such that
// N^M - N&M is maximum
static int getMaxDifference(int N)
{
    // Initialize variables
    int M = -1;
    int maxDiff = 0;
 
    // Iterate for all values < N
    for (int i = 0; i < N; i++)
    {
 
        // Find the difference between
        // Bitwise XOR and AND
        int diff = (N ^ i) - (N & i);
 
        // Check if new difference is
        // greater than previous maximum
        if (diff >= maxDiff)
        {
 
            // Update variables
            maxDiff = diff;
            M = i;
        }
    }
 
    // Return the answer
    return M;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given Number N
    int N = 6;
 
    // Function Call
    System.out.print(getMaxDifference(N));
}
}
 
// This code is contributed by Rohit_ranjan

Python3




# Python3 program for the above approach
 
# Function to return M<N such that
# N^M - N&M is maximum
def getMaxDifference(N):
     
    # Initialize variables
    M = -1;
    maxDiff = 0;
 
    # Iterate for all values < N
    for i in range(N):
 
        # Find the difference between
        # Bitwise XOR and AND
        diff = (N ^ i) - (N & i);
 
        # Check if new difference is
        # greater than previous maximum
        if (diff >= maxDiff):
             
            # Update variables
            maxDiff = diff;
            M = i;
 
    # Return the answer
    return M;
 
# Driver Code
if __name__ == '__main__':
     
    # Given number N
    N = 6;
 
    # Function call
    print(getMaxDifference(N));
 
# This code is contributed by amal kumar choubey

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to return M<N such that
// N^M - N&M is maximum
static int getMaxDifference(int N)
{
     
    // Initialize variables
    int M = -1;
    int maxDiff = 0;
 
    // Iterate for all values < N
    for(int i = 0; i < N; i++)
    {
         
        // Find the difference between
        // Bitwise XOR and AND
        int diff = (N ^ i) - (N & i);
 
        // Check if new difference is
        // greater than previous maximum
        if (diff >= maxDiff)
        {
             
            // Update variables
            maxDiff = diff;
            M = i;
        }
    }
 
    // Return the answer
    return M;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given number N
    int N = 6;
 
    // Function call
    Console.Write(getMaxDifference(N));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// javascript program for the above approach   
// Function to return M<N such that
    // N^M - N&M is maximum
    function getMaxDifference(N) {
        // Initialize variables
        var M = -1;
        var maxDiff = 0;
 
        // Iterate for all values < N
        for (i = 0; i < N; i++) {
 
            // Find the difference between
            // Bitwise XOR and AND
            var diff = (N ^ i) - (N & i);
 
            // Check if new difference is
            // greater than previous maximum
            if (diff >= maxDiff) {
 
                // Update variables
                maxDiff = diff;
                M = i;
            }
        }
 
        // Return the answer
        return M;
    }
 
    // Driver Code
     
        // Given Number N
        var N = 6;
 
        // Function Call
        document.write(getMaxDifference(N));
 
// This code contributed by aashish1995
</script>
Output: 
1

 

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

Efficient Approach: The idea is to observe that the difference between Bitwise XOR and Bitwise AND is maximum if Bitwise AND of the two numbers is the minimum possible number and the minimum possible number is 0
The Bitwise AND between the two numbers is zero if and only if they complement each other. Therefore, the possible value of M must be the complement of the given number N.
Below is the implementation of the above approach:
 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to flip all bits of N
int findM(int N)
{
    int M = 0;
 
    // Finding most significant bit of N
    int MSB = (int)log2(N);
 
    // Calculating required number
    for (int i = 0; i < MSB; i++) {
 
        if (!(N & (1 << i)))
            M += (1 << i);
    }
 
    // Return the answer
    return M;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 6;
 
    // Function Call
    cout << findM(N);
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to flip all bits of N
static int findM(int N)
{
    int M = 0;
 
    // Finding most significant bit of N
    int MSB = (int)Math.log(N);
 
    // Calculating required number
    for(int i = 0; i < MSB; i++)
    {
        if ((N & (1 << i)) == 0)
            M += (1 << i);
    }
 
    // Return the answer
    return M;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number
    int N = 6;
 
    // Function call
    System.out.print(findM(N));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program for the above approach
import math
 
# Function to flip all bits of N
def findM(N):
     
    M = 0;
 
    # Finding most significant bit of N
    MSB = int(math.log(N));
 
    # Calculating required number
    for i in range(MSB):
        if ((N & (1 << i)) == 0):
            M += (1 << i);
     
    # Return the answer
    return M;
 
# Driver Code
if __name__ == '__main__':
 
    # Given number
    N = 6;
 
    # Function call
    print(findM(N));
 
# This code is contributed by Amit Katiyar

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to flip all bits of N
static int findM(int N)
{
    int M = 0;
 
    // Finding most significant bit of N
    int MSB = (int)Math.Log(N);
 
    // Calculating required number
    for(int i = 0; i < MSB; i++)
    {
        if ((N & (1 << i)) == 0)
            M += (1 << i);
    }
 
    // Return the answer
    return M;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given number
    int N = 6;
 
    // Function call
    Console.Write(findM(N));
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
// javascript program for the above approach   
// Function to flip all bits of N
    function findM(N) {
        var M = 0;
 
        // Finding most significant bit of N
        var MSB = parseInt( Math.log(N));
 
        // Calculating required number
        for (i = 0; i < MSB; i++) {
            if ((N & (1 << i)) == 0)
                M += (1 << i);
        }
 
        // Return the answer
        return M;
    }
 
    // Driver Code
     
        // Given number
        var N = 6;
 
        // Function call
        document.write(findM(N));
 
// This code is contributed by Rajput-Ji
</script>
Output: 
1

 

Time Complexity: O(log2N) 
Auxiliary Space: O(1)
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :