Group all co-prime numbers from 1 to N

• Difficulty Level : Expert
• Last Updated : 08 Nov, 2021

Given an integer N, the task is to group numbers such that each group is mutually co-prime together with the total grouping is minimum.

Examples:

Input: N = 8
Output:
1 2 3
4 5
6 7
8

Input: N = 5
Output:
1 2 3
4 5

Approach: The key observation in this problem is two consecutive numbers are always co-prime. That is GCD(a, a+1) = 1. Another important observation is even numbers can’t be listed in one group. Because they will lead to the greatest common divisor of 2. Therefore, every consecutive even and odd numbers can be grouped into one group and 1 can be in any group because the greatest common divisor of numbers with 1 is always 1.

Below is the implementation of the above approach :

C++

 // C++ implementation to group// mutually coprime numbers into// one group with minimum group possible#includeusing namespace std; // Function to group the mutually// co-prime numbers into one groupvoid mutually_coprime(int n){    if (n <= 3)    {                 // Loop for the numbers less        // than the 4        for(int j = 1; j <= n; j++)        {            cout << j << " ";        }        cout << "\n";    }    else    {                 // Integers 1, 2 and 3 can be        // grouped into one group        cout << "1 2 3\n";                 for(int j = 4; j < n; j += 2)        {                         // Consecutive even and            // odd numbers            cout << j << " " << j + 1 << "\n";        }        if(n % 2 == 0)            cout << n << "\n";    }} // Driver Code        int main(){    int n = 9;         // Function call    mutually_coprime(n);} // This code is contributed by yatinagg

Java

 // Java implementation to group// mutually coprime numbers into// one group with minimum group possibleclass GFG{     // Function to group the mutually// co-prime numbers into one groupstatic void mutually_coprime(int n){    if (n <= 3)    {                 // Loop for the numbers less        // than the 4        for(int j = 1; j < n + 1; j++)           System.out.print(j + " ");        System.out.println();    }    else    {                 // Integers 1, 2 and 3 can be        // grouped into one group        System.out.println("1 2 3");        for(int j = 4; j < n; j += 2)        {            // Consecutive even and           // odd numbers           System.out.println(j + " " + (j + 1));           if (n % 2 == 0)           System.out.println(n);        }    }} // Driver Codepublic static void main(String[] args){    int n = 9;     // Function Call    mutually_coprime(n);}} // This code is contributed by sapnasingh4991

Python3

 # Python3 implementation to group# mutually coprime numbers into# one group with minimum group possible # Function to group the mutually# co-prime numbers into one groupdef mutually_coprime (n):       if ( n <= 3):        # Loop for the numbers less        # than the 4        for j in range (1, n + 1):            print (j, end =" ")        print ()    else:        # Integers 1, 2 and 3 can be        # grouped into one group        print (1, 2, 3)        for j in range ( 4, n, 2 ):                         # Consecutive even and            # odd numbers            print (j, ( j + 1 ))        if(n % 2 == 0):                    print (n) # Driver Code           if __name__ == "__main__":    n = 9         # Function Call    mutually_coprime (n)

C#

 // C# implementation to group// mutually coprime numbers into// one group with minimum group possibleusing System; class GFG{     // Function to group the mutually// co-prime numbers into one groupstatic void mutually_coprime(int n){    if (n <= 3)    {                 // Loop for the numbers less        // than the 4        for(int j = 1; j < n + 1; j++)           Console.Write(j + " ");                    Console.WriteLine();    }    else    {                 // ints 1, 2 and 3 can be        // grouped into one group        Console.WriteLine("1 2 3");        for(int j = 4; j < n; j += 2)        {           // Consecutive even and           // odd numbers           Console.WriteLine(j + " " + (j + 1));                        if (n % 2 == 0)               Console.WriteLine(n);        }    }} // Driver Codepublic static void Main(String[] args){    int n = 9;     // Function Call    mutually_coprime(n);}}// This code is contributed by sapnasingh4991

Javascript



Output:

1 2 3
4 5
6 7
8 9

Time Complexity: O(n)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up