Remainder Evaluation - Coding

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

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

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

Reffered: https://www.geeksforgeeks.org

Mathematical

Related
Sum of all N elements of the given series having X as its first integer
Minimum steps to reach Kaprekar's Constant
Check Whether a Number is an Anti Prime Number(Highly Composite Number)
Maximize value of given expression by choosing pair from Array
Minimize steps to make given two number equal by adding LSB

Type:	Geek
Category:	Coding
Sub Category:	Tutorial
Uploaded by:	Admin
Views:	18