Find a positive number M such that gcd(N^M, N&M) is maximum

Last Updated : 16 Jun, 2022

Given a number N, the task is to find a positive number M such that gcd(N^M, N&M) is the maximum possible and M < N. The task is to print the maximum gcd thus obtained.
Examples:

Input: N = 5 
Output: 7 
gcd(2^5, 2&5) = 7 

Input: N = 15 
Output: 5

Approach: There are two cases that need to be solved to get the maximum gcd possible.

If a minimum of one bit is not set in the number, then M will be a number whose bits are flipped at every position of N. And after that get the maximum gcd.
If all bits are set, then the answer will the maximum factor of that number except the number itself.

Below is the implementation of the above approach:

C++

// C++ program to implement above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to count
// the number of unset bits
int countBits(int n)
{
    // Base case
    if (n == 0)
        return 0;
 
    // unset bit count
    else
        return !(n & 1) + countBits(n >> 1);
}
 
// Function to return the max gcd
int maxGcd(int n)
{
 
    // If no unset bits
    if (countBits(n) == 0) {
        // Find the maximum factor
        for (int i = 2; i * i <= n; i++) {
            // Highest factor
            if (n % i == 0) {
                return n / i;
            }
        }
    }
    else {
 
        int val = 0;
        int power = 1;
        int dupn = n;
 
        // Find the flipped bit number
        while (n) {
            // If bit is not set
            if (!(n & 1)) {
                val += power;
            }
 
            // Next power of 2
            power = power * 2;
 
            // Right shift the number
            n = n >> 1;
        }
 
        // Return the answer
        return __gcd(val ^ dupn, val & dupn);
    }
 
    // If a prime number
    return 1;
}
 
// Driver Code
int main()
{
    // First example
    int n = 5;
    cout << maxGcd(n) << endl;
 
    // Second example
    n = 15;
    cout << maxGcd(n) << endl;
 
    return 0;
}

Java

// Java program to implement above approach
 
class GFG
{
 
static int __gcd(int a,int b)
{
    if(b == 0)
    return a;
    return __gcd(b, a % b);
}
 
// Recursive function to count
// the number of unset bits
static int countBits(int n)
{
    // Base case
    if (n == 0)
        return 0;
 
    // unset bit count
    else
        return ((n & 1) == 0 ? 1 : 0) + countBits(n >> 1);
}
 
// Function to return the max gcd
static int maxGcd(int n)
{
 
    // If no unset bits
    if (countBits(n) == 0)
    {
        // Find the maximum factor
        for (int i = 2; i * i <= n; i++) 
        {
            // Highest factor
            if (n % i == 0)
            {
                return n / i;
            }
        }
    }
    else
    {
 
        int val = 0;
        int power = 1;
        int dupn = n;
 
        // Find the flipped bit number
        while (n > 0) 
        {
            // If bit is not set
            if ((n & 1) == 0) 
            {
                val += power;
            }
 
            // Next power of 2
            power = power * 2;
 
            // Right shift the number
            n = n >> 1;
        }
 
        // Return the answer
        return __gcd(val ^ dupn, val & dupn);
    }
 
    // If a prime number
    return 1;
}
 
// Driver Code
public static void main(String[] args)
{
    // First example
    int n = 5;
    System.out.println(maxGcd(n));
 
    // Second example
    n = 15;
    System.out.println(maxGcd(n));
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python 3 program to implement 
# above approach
from math import gcd, sqrt
 
# Recursive function to count
# the number of unset bits
def countBits(n):
     
    # Base case
    if (n == 0):
        return 0
 
    # unset bit count
    else:
        return (((n & 1) == 0) +
                  countBits(n >> 1))
 
# Function to return the max gcd
def maxGcd(n):
     
    # If no unset bits
    if (countBits(n) == 0):
         
        # Find the maximum factor
        for i in range(2, int(sqrt(n)) + 1):
             
            # Highest factor
            if (n % i == 0):
                return int(n / i)
         
    else:
        val = 0
        power = 1
        dupn = n
 
        # Find the flipped bit number
        while (n):
             
            # If bit is not set
            if ((n & 1) == 0):
                val += power
 
            # Next power of 2
            power = power * 2
 
            # Right shift the number
            n = n >> 1
         
        # Return the answer
        return gcd(val ^ dupn, val & dupn)
 
    # If a prime number
    return 1
 
# Driver Code
if __name__ == '__main__':
     
    # First example
    n = 5
    print(maxGcd(n))
 
    # Second example
    n = 15
    print(maxGcd(n))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement above approach
using System;
class GFG
{
 
static int __gcd(int a,int b)
{
    if(b == 0)
    return a;
    return __gcd(b, a % b);
}
// Recursive function to count
// the number of unset bits
static int countBits(int n)
{
    // Base case
    if (n == 0)
        return 0;
 
    // unset bit count
    else
        return ((n & 1) == 0 ? 1 : 0) + countBits(n >> 1);
}
 
// Function to return the max gcd
static int maxGcd(int n)
{
 
    // If no unset bits
    if (countBits(n) == 0)
    {
        // Find the maximum factor
        for (int i = 2; i * i <= n; i++) 
        {
            // Highest factor
            if (n % i == 0)
            {
                return n / i;
            }
        }
    }
    else
    {
 
        int val = 0;
        int power = 1;
        int dupn = n;
 
        // Find the flipped bit number
        while (n > 0) 
        {
            // If bit is not set
            if ((n & 1) == 0) 
            {
                val += power;
            }
 
            // Next power of 2
            power = power * 2;
 
            // Right shift the number
            n = n >> 1;
        }
 
        // Return the answer
        return __gcd(val ^ dupn, val & dupn);
    }
 
    // If a prime number
    return 1;
}
 
// Driver Code
static void Main()
{
    // First example
    int n = 5;
    Console.WriteLine(maxGcd(n));
 
    // Second example
    n = 15;
    Console.WriteLine(maxGcd(n));
}
}
 
// This code is contributed by mits

PHP

<?php
// PHP program to implement above approach 
 
// Recursive function to count 
// the number of unset bits 
function countBits($n) 
{
    // Base case 
    if ($n == 0) 
        return 0; 
 
    // unset bit count 
    else
        return !($n & 1) + 
                 countBits($n >> 1); 
} 
 
// Recursive function to return
// gcd of a and b 
function gcd($a, $b) 
{ 
 
    // Everything divides 0 
    if ($a == 0) 
        return $b; 
    if ($b == 0) 
        return $a; 
 
    // base case 
    if($a == $b) 
        return $a ; 
     
    // a is greater 
    if($a > $b) 
        return gcd($a - $b , $b ); 
 
    return gcd($a , $b - $a ); 
} 
 
// Function to return the max gcd 
function maxGcd($n) 
{ 
 
    // If no unset bits 
    if (countBits($n) == 0) 
    { 
         
        // Find the maximum factor 
        for ($i = 2; $i * $i <= $n; $i++) 
        { 
             
            // Highest factor 
            if ($n % $i == 0) 
            { 
                return floor($n / $i); 
            } 
        } 
    } 
    else
    { 
        $val = 0; 
        $power = 1; 
        $dupn = $n; 
 
        // Find the flipped bit number 
        while ($n) 
        { 
             
            // If bit is not set 
            if (!($n & 1))
            { 
                $val += $power; 
            } 
 
            // Next power of 2 
            $power = $power * 2; 
 
            // Right shift the number 
            $n = $n >> 1; 
        } 
 
        // Return the answer 
        return gcd($val ^ $dupn, $val & $dupn); 
    } 
 
    // If a prime number 
    return 1; 
} 
 
// Driver Code 
 
// First example 
$n = 5; 
echo maxGcd($n), "\n"; 
 
// Second example 
$n = 15; 
echo maxGcd($n), "\n"; 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
// javascript program to implement above approach
    function __gcd(a , b) {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }
 
    // Recursive function to count
    // the number of unset bits
    function countBits(n) {
        // Base case
        if (n == 0)
            return 0;
 
        // unset bit count
        else
            return ((n & 1) == 0 ? 1 : 0) + countBits(n >> 1);
    }
 
    // Function to return the max gcd
    function maxGcd(n) {
 
        // If no unset bits
        if (countBits(n) == 0) {
            // Find the maximum factor
            for (i = 2; i * i <= n; i++) {
                // Highest factor
                if (n % i == 0) {
                    return n / i;
                }
            }
        } else {
 
            var val = 0;
            var power = 1;
            var dupn = n;
 
            // Find the flipped bit number
            while (n > 0) {
                // If bit is not set
                if ((n & 1) == 0) {
                    val += power;
                }
 
                // Next power of 2
                power = power * 2;
 
                // Right shift the number
                n = n >> 1;
            }
 
            // Return the answer
            return __gcd(val ^ dupn, val & dupn);
        }
 
        // If a prime number
        return 1;
    }
 
    // Driver Code
     
        // First example
        var n = 5;
        document.write(maxGcd(n)+"<br/>");
 
        // Second example
        n = 15;
        document.write(maxGcd(n));
 
// This code contributed by aashish1995
</script>

Output:

7
5

Time Complexity: O(sqrt(N) + logN), as we are using a loop to traverse sqrt(N) times, as the condition is i*i<=N, which is equivalent to i<=sqrt(N).

Auxiliary Space: O(logN), due to recursive stack space.

Suggest improvement

Check whether the exchange is possible or not

Minimum index i such that all the elements from index i to given index are equal

Share your thoughts in the comments