Largest number less than or equal to N/2 which is coprime to N

Given a number N, the task is to find the largest positive integer less than or equal to N/2 which is coprime to N. 
Note: Two number A and B are considered to coprime if gcd(A, B) = 1. It is also given that 2 < N < 10^18.
Examples:

Input: N = 50
Output: 23
GCD(50, 23) = 1 

Input: N = 100
Output: 49

Naive Approach: Start from N/2 and find the number smaller than or equal to N/2 which is coprime to N. 
Below is the implementation of the above approach: 

C++

// C++ implementation of the above approacdh
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to calculate gcd of two number
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

// Function to check if two numbers are coprime or not
bool coPrime(ll n1, ll n2)
{
    // two numbers are coprime if their gcd is 1
    if (gcd(n1, n2) == 1)
        return true;
    else
        return false;
}

// Function to find largest integer less
// than or equal to N/2 and coprime with N
ll largestCoprime(ll N)
{
    ll half = floor(N / 2);

    // Check one by one all numbers
    // less than or equal to N/2
    while (coPrime(N, half) == false)
        half--;

    return half;
}

// Driver code
int main()
{

    ll n = 50;
    cout << largestCoprime(n);

    return 0;
}

Java

// Java implementation of the above approacdh
import java.util.*;

class GFG
{

// Function to calculate gcd of two number
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

// Function to check if two
// numbers are coprime or not
static boolean coPrime(int n1, int n2)
{
    // two numbers are coprime
    // if their gcd is 1
    if (gcd(n1, n2) == 1)
        return true;
    else
        return false;
}

// Function to find largest integer less
// than or equal to N/2 and coprime with N
static int largestCoprime(int N)
{
    int half = (int)(N / 2);

    // Check one by one all numbers
    // less than or equal to N/2
    while (coPrime(N, half) == false)
        half--;

    return half;
}

// Driver code
public static void main(String args[])
{
    int n = 50;
    System.out.println(largestCoprime(n));
}
}

// This code is contributed by
// Surendra_Gangwar

Python3

# Python3 implementation of the above approacdh
import math as mt

# Function to calculate gcd of two number
def gcd( a, b):

    if (b == 0):
        return a
    else:
        return gcd(b, a % b)


# Function to check if two numbers are coprime or not
def coPrime( n1, n2):

    # two numbers are coprime if their gcd is 1
    if (gcd(n1, n2) == 1):
        return True
    else:
        return False


# Function to find largest integer less
# than or equal to N/2 and coprime with N
def largestCoprime( N):

    half = mt.floor(N / 2)

    # Check one by one a numbers
    # less than or equal to N/2
    while (coPrime(N, half) == False):
        half -= 1

    return half


# Driver code

n = 50
print( largestCoprime(n))

#This code is contributed by Mohit kumar 29

C#

// C# implementation of the above approacdh
using System;

class GFG
{

// Function to calculate gcd of two number
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

// Function to check if two
// numbers are coprime or not
static bool coPrime(int n1, int n2)
{
    // two numbers are coprime
    // if their gcd is 1
    if (gcd(n1, n2) == 1)
        return true;
    else
        return false;
}

// Function to find largest integer less
// than or equal to N/2 and coprime with N
static int largestCoprime(int N)
{
    int half = (int)(N / 2);

    // Check one by one all numbers
    // less than or equal to N/2
    while (coPrime(N, half) == false)
        half--;

    return half;
}

// Driver code
static void Main()
{
    int n = 50;
    Console.WriteLine(largestCoprime(n));
}
}

// This code is contributed by chandan_jnu

PHP

<?php
// PHP implementation of the above approach

// Function to calculate gcd of two number
function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    else
        return gcd($b, $a % $b);
}

// Function to check if two numbers
// are coprime or not
function coPrime($n1, $n2)
{
    // two numbers are coprime if
    // their gcd is 1
    if (gcd($n1, $n2) == 1)
        return true;
    else
        return false;
}

// Function to find largest integer less
// than or equal to N/2 and coprime with N
function largestCoprime($N)
{
    $half = floor($N / 2);

    // Check one by one all numbers
    // less than or equal to N/2
    while (coPrime($N, $half) == false)
        $half--;

    return $half;
}

// Driver code
$n = 50;
echo largestCoprime($n);

// This code is contributed
// by Akanksha Rai

Javascript

// Javascript implementation of the above approach

// Function to calculate gcd of two number
function gcd(a, b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

// Function to check if two numbers
// are coprime or not
function coPrime(n1, n2)
{
    // two numbers are coprime if
    // their gcd is 1
    if (gcd(n1, n2) == 1)
        return true;
    else
        return false;
}

// Function to find largest integer less
// than or equal to N/2 and coprime with N
function largestCoprime(N)
{
    let half = Math.floor(N / 2);

    // Check one by one all numbers
    // less than or equal to N/2
    while (coPrime(N, half) == false)
        half--;

    return half;
}

// Driver code
let n = 50;
document.write(largestCoprime(n));

// This code is contributed
// by gfgking

Output:

23

Time Complexity : O(nlogn)

Auxiliary Space: O(logn)

Efficient Approach: To observe the pattern:

• If the given number is odd, the largest coprime number will be (N-1)/2.
• If the given number is divisible by 4, the largest coprime number will be (N)/2 – 1.
• If the given number is divisible by 2, the largest coprime number will be (N)/2 – 2.

Note: There is a special case 6, for which greatest coprime number less than N / 2 will be 1.
Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find largest integer less than
// or equal to N/2 and is coprime with N
long long largestCoprime(long long N)
{
    // Handle the case for N = 6
    if (N == 6)
        return 1;

    else if (N % 4 == 0)
        return (N / 2) - 1;

    else if (N % 2 == 0)
        return (N / 2) - 2;

    else
        return ((N - 1) / 2);
}

// Driver code
int main()
{

    long long int n = 50;
    cout << largestCoprime(n) << endl;

    return 0;
}

Java

// Java implementation of the above approach
class GfG
{

    // Function to find largest integer less than
    // or equal to N/2 and is coprime with N
    static int largestCoprime(int N)
    {

        // Handle the case for N = 6
        if (N == 6)
            return 1;

        else if (N % 4 == 0)
            return (N / 2) - 1;

        else if (N % 2 == 0)
            return (N / 2) - 2;

        else
            return ((N - 1) / 2);
    }

    // Driver code
    public static void main(String []args)
    {
        int n = 50;
        System.out.println(largestCoprime(n));
    }
}

// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the above approach

# Function to find largest integer less than
# or equal to N/2 and is coprime with N
def largestCoprime(N):

    # Handle the case for N = 6
    if N == 6:
        return 1

    elif N % 4 == 0:
        return N // 2 - 1

    elif N % 2 == 0:
        return N // 2 - 2

    else:
        return (N - 1) // 2

# Driver code
if __name__ == "__main__":

    n = 50
    print(largestCoprime(n))

# This code is contributed by Rituraj Jain

C#

// C# implementation of the above approach
using System;

class GfG
{

    // Function to find largest
    // integer less than or equal
    // to N/2 and is coprime with N
    static int largestCoprime(int N)
    {

        // Handle the case for N = 6
        if (N == 6)
            return 1;

        else if (N % 4 == 0)
            return (N / 2) - 1;

        else if (N % 2 == 0)
            return (N / 2) - 2;

        else
            return ((N - 1) / 2);
    }

    // Driver code
    public static void Main()
    {
        int n = 50;
        Console.WriteLine(largestCoprime(n));
    }
}

// This code is contributed by Ryuga

PHP

<?php
// PHP implementation of the above approach

// Function to find largest integer less than
// or equal to N/2 and is coprime with N
function largestCoprime($N)
{
    // Handle the case for N = 6
    if ($N == 6)
        return 1;

    else if ($N % 4 == 0)
        return ($N / 2) - 1;

    else if ($N % 2 == 0)
        return ($N / 2) - 2;

    else
        return (($N - 1) / 2);
}

// Driver code
$n = 50;
echo largestCoprime($n);

// This code is contributed by
// chandan_jnu
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to find largest integer less than
// or equal to N/2 and is coprime with N
function largestCoprime(N)
{

    // Handle the case for N = 6
    if (N == 6)
        return 1;

    else if (N % 4 == 0)
        return (N / 2) - 1;

    else if (N % 2 == 0)
        return (N / 2) - 2;

    else
        return ((N - 1) / 2);
}

// Driver Code
var n = 50;

document.write(largestCoprime(n));

// This code is contributed by Khushboogoyal499

</script>

Output:

23

Time Complexity : O(1), since there are only basic arithmetic operations that take constant time.

Auxiliary Space: O(1), since no extra space has been required.