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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return s % nint 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 codeint main(){ string s = "81"; int n = s.length(); cout << findMod4(s, n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return s % nstatic 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 codepublic 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 % ndef 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 codeif __name__ == '__main__': s = "81" n = len(s) print(findMod4(s, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return s % nstatic 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 codepublic 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 % nfunction 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 % nfunction 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 codevar 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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