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++ 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 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 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# 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. |
<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>
|
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++ 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;
} |
/*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 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__ |
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__. |
<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> |
2
Time Complexity: O(1)
Auxiliary Space: O(1)