HCF of Two Number in C++

September 29, 2022

HCF of Two Number

Here we will discuss how to find the HCF of Two Number (also known as GCD) using C++ programming language.

HCF ( Highest Common Factor ) of two numbers is the largest positive integer that can divide both the numbers

GCD or HCF of two numbers
While loop in C

We will learn

  • Method 1: Linear Quest to find HCF
  • Method 2: Euclidean Algorithm: Repeated Subtraction
  • Method 3: Recursive Euclidean Algorithm: Repeated Subtraction
  • Method 4: Modulo Recursive Euclidean Algorithm: Repeated Subtraction
  • Method 5: Handling Negative Numbers in HCF

Method 1 : Linear Quest

Algorithm

  • Initialize HCF = 1
  • Run a loop in the iteration of (i) between [1, min(num1, num2)]
  • Note down the highest number that divides both num1 & num2
  • If i satisfies (num1 % i == 0 && num2 % i == 0) then new value of HCF is i
  • Print value of HCF

Method 1 : C++ Code

Run 
#include<iostream>
using namespace std;

int main()
{
    int num1 = 36, num2 = 60, hcf = 1;

    for(int i = 1; i <= num1 || i <= num2; i++)
    {
        if(num1 % i == 0 && num2 % i == 0)
            hcf = i;
    }

    cout<<"HCF of "<<num1<<" and "<<num2<<" is "<<hcf;

    return 0;
}

Output

HCF of 36 and 60 is 12

Lowest Common Multiple (LCM)

 

Binary to Decimal to conversion

Octal to Decimal conversion

Hexadecimal to Decimal conversion

Method 2 : Repeated Subtraction

Algorithm

  • Run a while loop until num1 is not equals to num2
  • If num1>num2 then num1 = num1 – num2
  • Else num2 = num2 – num1
  • After the loop ends both num1 & num2 stores HCF

Method 2 : C++ Code

Run 
#include<iostream>
using namespace std;

int main()
{
    int num1 = 36, num2 = 60;
    int a = num1, b = num2;

    while (num1 != num2)
    {
        if (num1 > num2)
            num1 -= num2;
        else
            num2 -= num1;
    }

    cout<<"HCF of "<<a<<" and "<<b<<" is "<<num1;

    return 0;
}

Output

HCF of 36 and 60 is 12

Method 3 : Repeated Subtraction using Recursion

Algorithm

  • Checked whether any of the input is 0 then return sum of both numbers
  • If both input are equal return any of the two numbers
  • If num1 is greater than the num2 then Recursively call findHCF(num1 – num2, num2)
  • Else Recursively call findHCF(num1, num2-num1)

Method 3 : C++ Code

Run 
// C++ program to find HCF of two numbers
#include<iostream>

using namespace std;

int findHCF(int, int);

int main()
{
    int num1 = 36, num2 = 60;

    cout<< "HCF of "<< num1<< " and "<< num2 <<" is "<< findHCF(num1, num2); 
    
    return 0; 
} 

// Recursive function to return HCF of a and b int 
int findHCF(int num1, int num2) 
{ 
    // Everything divides 0 
    if(num1==0 || num2==0) 
    { 
        return ( num1 + num2 ); 
    } 

    // base case 
    if(num1==num2) 
    { 
         return num1; 
    } 

    // num1 > num2
    if(num1>num2)
    {
        return findHCF(num1 - num2, num2);
    }
    else
    {
        return findHCF(num1, num2 - num1);
    }
}

Output

HCF of 36 and 60 is 12

Method 4 : Repeated Subtraction with Modulo Operator using Recursion

Algorithm

  • If b is equals to 0 return a
  • Else recursively call the function for value b, a%b and return 

Method 4 : C++ Code

Run 
#include<iostream>
using namespace std;

// This method improves complexity of repeated substraction
// By efficient use of modulo operator in euclidean algorithm
int getHCF(int a, int b)
{
    return b == 0 ? a : getHCF(b, a % b); 
}

int main()
{
    int num1 = 36, num2 = 60;

    cout<<"HCF of "<<num1<<" and "<<num2<<" is "<<getHCF(num1, num2);

    return 0;
}
// Time Complexity: log(max(a,b))

Output

HCF of 36 and 60 is 12

Method 5 : Handling Negative Numbers in HCF

Algorithm

If any of the number is negative then convert it to positive by multiplying it with -1 as according to the proper definition HCF of two numbers can never be negative.

  • If b is equals to 0 return a
  • Else recursively call the function for value b, a%b and return 

Method 5 : C++ Code

Run 
#include<iostream>
using namespace std;

// This method improves complexity of repeated substraction
// By efficient use of modulo operator in euclidean algorithm
int getHCF(int a, int b)
{
    return b == 0 ? a : getHCF(b, a % b); 
}

int main()
{
    int num1 = -36, num2 = 60;

    // if user enters negative number, we just changing it to positive
    // By definition HCF is highest positive number that divides both numbers
    // -36 & 60 : HCF = 12 (as highest num that divides both)
    // -36 & -60 : HCF = 16 (as highest num that divides both)
    num1 = (num1 > 0) ? num1 : -num1;
    num2 = (num2 > 0) ? num2 : -num2;
    
    cout<<"HCF of "<<num1<<" and "<<num2<<" is "<<getHCF(num1, num2);
    
    return 0;
}
// Time Complexity: log(max(a,b))

Output

HCF of -36 and 60 is 12

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While loop in C