Maximum possible GCD for a pair of integers with sum N

Given an integer N, the task is to find the maximum possible GCD of a pair of integers such that their sum is N.

Examples :

Input: N = 30 
Output: 15 
Explanation: GCD of (15, 15) is 15, which is the maximum possible GCD

Input: N = 33 
Output: 11 
Explanation: GCD of (11, 22) is 11, which is the maximum possible GCD

Naive Approach: 
The simplest approach to solve this problem is to calculate GCD for all pair of integers with sum N and find the maximum possible GCD among them. 
Time complexity: O(N²logN) 
Auxiliary Space: O(1)

Efficient Approach: 
Follow the steps given below to optimize the above approach:

• Iterate up to √N and find the largest proper factor of N.
• If N is prime, i.e. no factor could be obtained, print 1, as all pairs are co-prime.
• Otherwise, print the largest possible factor as the answer.

Below is the implementation of the above approach:

C++

// C++ Program to find the maximum
// possible GCD of any pair with
// sum N
#include <bits/stdc++.h>
using namespace std;

// Function to find the required
// GCD value
int maxGCD(int N)
{
    for (int i = 2; i * i <= N; i++) {

        // If i is a factor of N
        if (N % i == 0) {

            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}

// Driver Code
int main()
{
    int N = 33;
    cout << "Maximum Possible GCD value is : "
        << maxGCD(N) << endl;
    return 0;
}

Java

// Java program to find the maximum
// possible GCD of any pair with
// sum N
import java.io.*;
public class GFG{

// Function to find the required
// GCD value
static int maxGCD(int N)
{
    for(int i = 2; i * i <= N; i++)
    {

        // If i is a factor of N
        if (N % i == 0)
        {

            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}

// Driver Code
public static void main(String[] args)
{
    int N = 33;
    System.out.println("Maximum Possible GCD " +
                       "value is : " + maxGCD(N));
}
}

// This code is conhtributed by rutvik_56

Python3

# Python3 program to find the maximum
# possible GCD of any pair with
# sum N

# Function to find the required
# GCD value
def maxGCD(N):

    i = 2
    while(i * i <= N):

        # If i is a factor of N
        if (N % i == 0):

            # Return the largest
            # factor possible
            return N // i

        i += 1

    # If N is a prime number
    return 1

# Driver Code
N = 33

print("Maximum Possible GCD value is : ",
       maxGCD(N))

# This code is contributed by code_hunt

C#

// C# program to find the maximum
// possible GCD of any pair with
// sum N
using System;

class GFG{

// Function to find the required
// GCD value
static int maxGCD(int N)
{
    for(int i = 2; i * i <= N; i++)
    {

        // If i is a factor of N
        if (N % i == 0)
        {

            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}

// Driver Code
public static void Main(String[] args)
{
    int N = 33;
    Console.WriteLine("Maximum Possible GCD " +
                      "value is : " + maxGCD(N));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program to find the maximum
// possible GCD of any pair with
// sum N

// Function to find the required
// GCD value
function maxGCD(N)
{
    for(var i = 2; i * i <= N; i++)
    {

        // If i is a factor of N
        if (N % i == 0)
        {

            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}

// Driver code
var N = 33;
document.write("Maximum Possible GCD " +
               "value is :" + maxGCD(N));

// This code is contributed by Ankita saini.

</script>

Output:

Maximum Possible GCD value is : 11

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