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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 ``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'``;` `    ``// 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'``;` `    ``// 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

 `

## Javascript

 ``
Output:

`1`

Time Complexity: O(1)

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