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Find N % 4 (Remainder with 4) for a large value of N
  • Difficulty Level : Easy
  • Last Updated : 19 Mar, 2021

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:
 

 

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++




// 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;
}

Java




// 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.

Python3




# 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

C#




// 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.

PHP




<?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.

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return s % n
function findMod4(s, n)
{
 
    // To store the number formed by
    // the last two digits
    var k=0;
 
    // 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
var s = "81";
var n = s.length;
document.write(findMod4(s, n));
 
// This code is contributed by nood2000.
</script>
Output: 



1

 

Time Complexity: O(1)
 

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