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

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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.


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>


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.


Output

1







Time Complexity: O(1)

Approach: Using Bitwise Operators

Steps:

  • The input number is read as a string.
  • The string is converted to an integer using a loop that iterates over each character of the string.
  • Updates the integer by multiplying it by 10 and adding the value of the current digit.
  • The remainder of the integer n divided by 4 is found using the bitwise AND operator & and the number 3, which is equal to 0b11 in binary.
  • This operation keeps only the two least significant bits of the integer, which represent the remainder when divided by 4.
  • Last, print the remainder.

Below is implementation of the above approach:

C++




#include <iostream>
#include <string>
using namespace std;
 
int main() {
    string str="81";
 
    // Convert the string to an integer
    unsigned long long n = 0;
    for (char c : str) {
        n = n * 10 + (c - '0');
    }
 
    // Find the remainder of n divided by 4 using bitwise AND
    int remainder = n & 3;
 
    cout << remainder << endl;
 
    return 0;
}


Java




public class Main {
    public static void main(String[] args) {
        String str = "81";
 
        // Convert the string to a long
        long n = 0;
        for (char c : str.toCharArray()) {
            n = n * 10 + (c - '0');
        }
 
        // Find the remainder of n divided by 4 using bitwise AND
        int remainder = (int) (n & 3);
 
        System.out.println(remainder);
    }
}


Python3




str = "81"
 
# Convert the string to an integer
n = 0
for c in str:
    n = n * 10 + int(c)
 
# Find the remainder of n divided by 4 using bitwise AND
remainder = n & 3
 
print(remainder)


C#




using System;
 
namespace ConsoleApp
{
    class GFG
    {
        static void Main(string[] args)
        {
            string str = "81";
 
            // Convert the string to a long integer
            ulong n = 0;
            foreach (char c in str)
            {
                n = n * 10 + (ulong)(c - '0');
            }
 
            // Find the remainder of n divided by 4 using bitwise AND
            int remainder = (int)(n & 3);
 
            Console.WriteLine(remainder);
        }
    }
}


Javascript




// Define the input string
let str = "81";
 
// Initialize a variable to store the converted number
let n = 0;
 
// Convert the string to a number
for (let i = 0; i < str.length; i++) {
    // Multiply the current result by 10 and add the numeric value of the character
    n = n * 10 + parseInt(str[i]);
}
 
// Find the remainder of n divided by 4 using bitwise AND
let remainder = n & 3;
 
// Print the remainder
console.log(remainder);


Output

1








Time Complexity: O(n), where n is the number of digits in the input string
Auxiliary Space: O(1) 



Last Updated : 14 Oct, 2023
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