Remainder Evaluation

Given two positive integers Num1 and Num2, the task is to find the remainder when Num1 is divided by Num2.

Examples:

Input:  Num1 = 11, Num2 = 3
Output: 2
Explanation:  3) 11 (3
                                – 9
                           ———
                                  2 -> Remainder
                          ———-

Input: Num1 = 15, Num2 = 3
Output: 0

Approach 1: The problem can be solved by using the modulus operator.

• Modulus operator returns the remainder, if we write a % b, it returns the remainder when a is divided by b where b != 0. If b = 0, then it gives Runtime Error,
  • Math error in C++, (Math error: Attempted to divide by Zero)
  • ZeroDivisionError in Python, [ZeroDivisionError: integer division or modulo by zero]
  • ArithmeticException in Java [ArithmeticException: / by zero]

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the remainder
// when Num1 is divided by Num2
int solve(int Num1, int Num2)
{
    return Num1 % Num2;
}

// Driver Code
int main()
{
    int Num1 = 11;
    int Num2 = 3;

    // Function Call
    cout << solve(Num1, Num2) << endl;
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;

class GFG {
    // Function to find the remainder
    // when Num1 is divided by Num2
    public static int solve(int Num1, int Num2)
    {
        return Num1 % Num2;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int Num1 = 11;
        int Num2 = 3;

        // Function Call
        System.out.println(solve(Num1, Num2));
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python3 code to implement the approach

# Function to find the remainder
# when Num1 is divided by Num2
def solve(Num1, Num2):
    return Num1 % Num2

# Driver Code
Num1 = 11
Num2 = 3

# Function Call
print(solve(Num1, Num2))

# This code is contributed by akashish__

C#

// C# program to implement
// the above approach
using System;
class GFG
{

    // Function to find the remainder
    // when Num1 is divided by Num2
    public static int solve(int Num1, int Num2)
    {
        return Num1 % Num2;
    }

// Driver Code
public static void Main()
{
    int Num1 = 11;
    int Num2 = 3;

    // Function Call
    Console.Write(solve(Num1, Num2));
}
}

// This code is contributed by sanjoy_62.

Javascript

<script>
        // JS code to implement the approach

        // Function to find the remainder
        // when Num1 is divided by Num2
        function solve(Num1, Num2) {
            return Num1 % Num2;
        }

        // Driver Code
        let Num1 = 11;
        let Num2 = 3;

        // Function Call
        document.write(solve(Num1, Num2));

// This code is contributed by lokeshpotta20.
    </script>

Output

2

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

Approach 2: Without using the modulus (%) operator

In this approach, we will consider Num2 as the divider and Num1 as the Dividend. so Quotient will be Num1 / Num2. then we will subtract (Quotient * Num2) from Num1, and this will be the Remainder.

Quotient = Num1 / Num2
Reminder = Num1 - (Quotient * Num2)

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the remainder when Num1 is divided by
// Num2 without using % operator
int solve(int Num1, int Num2)
{
    return Num1 - ((Num1 / Num2) * Num2);
}

// Driver Code
int main()
{
    int Num1 = 11;
    int Num2 = 3;

    // Function Call
    cout << solve(Num1, Num2) << endl;
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {

  // Function to find the remainder when Num1 is divided
  // by Num2 without using % operator
  static int solve(int Num1, int Num2)
  {
    return Num1 - ((Num1 / Num2) * Num2);
  }

  public static void main(String[] args)
  {
    int Num1 = 11;
    int Num2 = 3;

    // Function Call
    System.out.println(solve(Num1, Num2));
  }
}

// This code is contributed by aadityaburujwale.

Python3

# Python3 code to implement the above approach

# Function to find the remainder when Num1 is divided by
# Num2 without using % operator
def solve(Num1, Num2):
  return Num1 - (int(Num1 / Num2) * Num2);

# Driver Code
Num1 = 11
Num2 = 3

# Function Call
print(int(solve(Num1, Num2)))

# This code is contributed by akashish__

C#

using System;

public class GFG {

  // Function to find the remainder when Num1 is divided
  // by
  // Num2 without using % operator
  public static int solve(int Num1, int Num2)
  {
    return Num1 - ((Num1 / Num2) * Num2);
  }

  static public void Main()
  {

    int Num1 = 11;
    int Num2 = 3;

    // Function Call
    Console.WriteLine(solve(Num1, Num2));
  }
}

// This code is contributed by akashish__.

Javascript

<script>

// Function to find the remainder when Num1 is divided by
// Num2 without using % operator
function solve( Num1,Num2)
{
    return Num1 - (Math.floor(Num1 / Num2) * Num2);
}

// Driver Code
let Num1 = 11;
let Num2 = 3;

// Function Call
console.log(solve(Num1, Num2));

// This code is contributed by akashish__

</script>

Output

2

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