Find N % 4 (Remainder with 4) for a large value of N

Given a string str representing a large integer, the task is to find the result of N % 4.

Examples:

Input: N = 81
Output: 1

Input: N = 46234624362346435768440
Output: 0



Approach: The remainder of division by 4 is dependent on only the last 2 digits of a number, so instead of dividing N we divide only the last two digits of N and find the remainder.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return s % n
int findMod4(string s, int n)
{
  
    // To store the number formed by
    // the last two digits
    int k;
  
    // If it contains a single digit
    if (n == 1)
        k = s[0] - '0';
  
    // Take last 2 digits
    else
        k = (s[n - 2] - '0') * 10
            + s[n - 1] - '0';
  
    return (k % 4);
}
  
// Driver code
int main()
{
    string s = "81";
    int n = s.length();
    cout << findMod4(s, n);
  
    return 0;
}

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Java














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// Java implementation of the approach 


class GFG 


{ 


      


// Function to return s % n 


static int findMod4(String s, int n) 


{ 


  


    // To store the number formed by 


    // the last two digits 


    int k; 


  


    // If it contains a single digit 


    if (n == 1) 


        k = s.charAt(0) - '0'; 


  


    // Take last 2 digits 


    else


        k = (s.charAt(n - 2) - '0') * 10


            + s.charAt(n - 1) - '0'; 


  


    return (k % 4); 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    String s = "81"; 


    int n = s.length(); 


    System.out.println(findMod4(s, n)); 


} 


} 


  


// This code is contributed by Code_Mech. 



















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Python3














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# Python 3 implementation of the approach 


  


# Function to return s % n 


def findMod4(s, n): 


      


    # To store the number formed by 


    # the last two digits 


      


    # If it contains a single digit 


    if (n == 1): 


        k = ord(s[0]) - ord('0') 


  


    # Take last 2 digits 


    else: 


        k = ((ord(s[n - 2]) - ord('0')) * 10 + 


              ord(s[n - 1]) - ord('0')) 


  


    return (k % 4) 


  


# Driver code 


if __name__ == '__main__': 


    s = "81"


    n = len(s) 


    print(findMod4(s, n)) 


  


# This code is contributed by 


# Surendra_Gangwar 



















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C#














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// C# implementation of the approach 


using System; 


class GFG 


{ 


      


// Function to return s % n 


static int findMod4(string s, int n) 


{ 


  


    // To store the number formed by 


    // the last two digits 


    int k; 


  


    // If it contains a single digit 


    if (n == 1) 


        k = s[0] - '0'; 


  


    // Take last 2 digits 


    else


        k = (s[n - 2]- '0') * 10 


            + s[n - 1] - '0'; 


  


    return (k % 4); 


} 


  


// Driver code 


public static void Main() 


{ 


    string s = "81"; 


    int n = s.Length; 


    Console.WriteLine(findMod4(s, n)); 


} 


} 


  


// This code is contributed by Code_Mech. 



















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PHP














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<?php 


  


// PHP implementation of the approach 


// Function to return s % n 


function findMod4($s, $n) 


{ 


  


    // To store the number formed by 


    // the last two digits 


    $k; 


  


    // If it contains a single digit 


    if ($n == 1) 


        $k = $s[0] - '0'; 


  


    // Take last 2 digits 


    else


        $k = ($s[$n - 2] - '0') * 10 


            + $s[$n - 1] - '0'; 


  


    return ($k % 4); 


} 


  


// Driver code 


{ 


    $s = "81"; 


    $n = strlen($s); 


    echo(findMod4($s, $n)); 


} 


  


// This code is contributed by Code_Mech. 



















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Output:


1





Time Complexity: O(1)



          
          
          
          

            

    

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