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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