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

0 0 7 minutes read

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:

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:

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.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!