Check if a large number is divisible by 11 or not

Given a number, the task is to check if the number is divisible by 11 or not. The input number may be large and it may not be possible to store it even if we use long long int.

Examples: 

Input : n = 76945
Output : Yes

Input : n = 1234567589333892
Output : Yes

Input : n = 363588395960667043875487
Output : No

[Naive Approach]: Modulo Division Method

Checking given number is divisible by 11 or not by using the modulo division operator “%”.  

// C++ code to check whether
// the given number is divisible by 11 or not
#include <bits/stdc++.h>
using namespace std;

int main()
{
  
    // input
    long long n = 1234567589333892;

    // the above input can also be given as n=input() ->
    // taking input from user finding given number is
    // divisible by 11 or not
    if (n % 11 == 0)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

#include <stdio.h>

int main() {
    // input
    long long n = 1234567589333892;

    // the above input can also be given as n=input() ->
    // taking input from user finding given number is
    // divisible by 11 or not
    if (n % 11 == 0)
        printf("Yes\n");
    else
        printf("No\n");
    return 0;
}

public class Main {
    public static void main(String[] args) {
        // input
        long n = 1234567589333892L;

        // the above input can also be given as n=input() ->
        // taking input from user finding given number is
        // divisible by 11 or not
        if (n % 11 == 0)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

n = 1234567589333892

# the above input can also be given as n=input() ->
# taking input from user finding given number is
# divisible by 11 or not
if n % 11 == 0:
    print("Yes")
else:
    print("No")

using System;

class Program {
    static void Main() {
        // input
        long n = 1234567589333892;

        // the above input can also be given as n=input() ->
        // taking input from user finding given number is
        // divisible by 11 or not
        if (n % 11 == 0)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// input
let n = 1234567589333892;

// the above input can also be given as n=input() ->
// taking input from user finding given number is
// divisible by 11 or not
if (n % 11 === 0) {
    console.log("Yes");
} else {
    console.log("No");
}

Yes

Time Complexity: O(1) because it is performing constant operations
Auxiliary Space: O(1)

[Expected Approach] – Even-Odd Digit Sum for Large String Input

Since input number may be very large, we cannot use n % 11 to check if a number is divisible by 11 or not, especially in languages like C/C++. The idea is based on following fact.
A number is divisible by 11 if difference of following two is divisible by 11. 

1. Sum of digits at odd places.
2. Sum of digits at even places.

Illustration: 

For example, let us consider 76945
Sum of digits at odd places : 7 + 9 + 5, Sum of digits at even places : 6 + 4
Difference of two sums = 21 – 10 = 11. Since difference is divisible by 11, the number 7945 is divisible by 11.

How does this work? 

Let us consider 7694, we can write it as: 7694 = 7*1000 + 6*100 + 9*10 + 4
The proof is based on below observation:

• Remainder of 10ⁱ divided by 11 is 1 if i is even
• Remainder of 10ⁱ divided by 11 is -1 if i is odd

So the powers of 10 only result in values either 1 or -1. Remainder of “7*1000 + 6*100 + 9*10 + 4” divided by 11 can be written as :
7*(-1) + 6*1 + 9*(-1) + 4*1
The above expression is basically difference between sum of even digits and odd digits.

// C++ program to find if a number is divisible by
// 11 or not
#include<bits/stdc++.h>
using namespace std;

// Function to find that number divisible by 11 or not
int check(string str)
{
    int n = str.length();

    // Compute sum of even and odd digit
    // sums
    int oddDigSum = 0, evenDigSum = 0;
    for (int i=0; i<n; i++)
    {
        // When i is even, position of digit is odd
        if (i%2 == 0)
            oddDigSum += (str[i]-'0');
        else
            evenDigSum += (str[i]-'0');
    }

    // Check its difference is divisible by 11 or not
    return ((oddDigSum - evenDigSum) % 11 == 0);
}

// Driver code
int main()
{
    string str = "76945";
    check(str)?  cout << "Yes" : cout << "No ";
    return 0;
}

// Java program to find if a number is
// divisible by 11 or not
import java.io.*;
class IsDivisible
{
    // Function to find that number divisible by 11 or not
    static boolean check(String str)
    {
        int n = str.length();
     
        // Compute sum of even and odd digit
        // sums
        int oddDigSum = 0, evenDigSum = 0;
        for (int i=0; i<n; i++)
        {
            // When i is even, position of digit is odd
            if (i%2 == 0)
                oddDigSum += (str.charAt(i)-'0');
            else
                evenDigSum += (str.charAt(i)-'0');
        }
     
        // Check its difference is divisible by 11 or not
        return ((oddDigSum - evenDigSum) % 11 == 0);
    }
    
    // main function
    public static void main (String[] args) 
    {
        String str = "76945";
        if(check(str))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
} 

# Python 3 code program to find if a number 
# is divisible by 11 or not


# Function to find that number divisible by
#  11 or not
def check(st) :
    n = len(st) 

    # Compute sum of even and odd digit
    # sums
    oddDigSum = 0
    evenDigSum = 0
    for i in range(0,n) :
        # When i is even, position of digit is odd
        if (i % 2 == 0) :
            oddDigSum = oddDigSum + ((int)(st[i]))
        else:
            evenDigSum = evenDigSum + ((int)(st[i]))
    
    
    # Check its difference is divisible by 11 or not
    return ((oddDigSum - evenDigSum) % 11 == 0)

# Driver code
st = "76945"
if(check(st)) :
    print( "Yes")
else : 
    print("No ")
    
# This code is contributed by Nikita tiwari.

// C# program to find if a number is
// divisible by 11 or not
using System;

class GFG
{
    // Function to find that number 
    // divisible by 11 or not
    static bool check(string str)
    {
        int n = str.Length;
    
        // Compute sum of even and odd digit
        // sums
        int oddDigSum = 0, evenDigSum = 0;
        
        for (int i = 0; i < n; i++)
        {
            // When i is even, position of
            // digit is odd
            if (i % 2 == 0)
                oddDigSum += (str[i] - '0');
            else
                evenDigSum += (str[i] - '0');
        }
    
        // Check its difference is
        // divisible by 11 or not
        return ((oddDigSum - evenDigSum) 
                                % 11 == 0);
    }
    
    // main function
    public static void Main () 
    {
        String str = "76945";
        
        if(check(str))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
} 

// This code is contributed by vt_m.

// JavaScript program for the above approach

    // Function to find that number 
    // divisible by 11 or not
    function check(str)
    {
        let n = str.length;
    
        // Compute sum of even and odd digit
        // sums
        let oddDigSum = 0, evenDigSum = 0;
        
        for (let i = 0; i < n; i++)
        {
        
            // When i is even, position of
            // digit is odd
            if (i % 2 == 0)
                oddDigSum += (str[i] - '0');
            else
                evenDigSum += (str[i] - '0');
        }
    
        // Check its difference is
        // divisible by 11 or not
        return ((oddDigSum - evenDigSum) 
                                % 11 == 0);
    }
    
// Driver Code
    let str = "76945";
    if(check(str))
        console.log("Yes");
    else
        console.log("No");
        

Yes

Time Complexity: O(n), where n is the given number.
Auxiliary Space: O(1), as we are not using any extra space.