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# Program to find remainder when large number is divided by 11

• Difficulty Level : Medium
• Last Updated : 30 Mar, 2023

Given a number n, the task is to find the remainder when n is divided by 11. The input of number may be very large.

Examples:

```Input : str = 13589234356546756
Output : 6

Input : str = 3435346456547566345436457867978
Output : 4```

Since the given number can be very large, we can not use n % 11. There are some steps that needs to be used to find remainder:

```1. Store number in string.
2. Count length of number string.
3. Convert string character one by one
into digit and check if it's less than
11. Then continue for next character
otherwise take remainder and use
remainder for next number.
4. We get remainder.

Ex. str = "1345"
len = 4
rem = 3```

Implementation:

## C++

 `// CPP implementation to find remainder` `// when a large number is divided by 11` `#include ` `using` `namespace` `std;`   `// Function to return remainder` `int` `remainder(string str)` `{` `    ``// len is variable to store the` `    ``// length of number string.` `    ``int` `len = str.length();`   `    ``int` `num, rem = 0;`   `    ``// loop that find remainder` `    ``for` `(``int` `i = 0; i < len; i++) {` `        ``num = rem * 10 + (str[i] - ``'0'``);` `        ``rem = num % 11;` `    ``}`   `    ``return` `rem;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"3435346456547566345436457867978"``;` `    ``cout << remainder(str);` `    ``return` `0;` `}`

## Java

 `// JAVA implementation to find remainder` `// when a large number is divided by 11` `import` `java.io.*;`   `class` `GFG{` `    `  `    ``// Function to return remainder` `    ``static` `int` `remainder(String str)` `    ``{` `        ``// len is variable to store the` `        ``// length of number string.` `        ``int` `len = str.length();` `     `  `        ``int` `num, rem = ``0``;` `     `  `        ``// loop that find remainder` `        ``for` `(``int` `i = ``0``; i < len; i++) {` `            ``num = rem * ``10` `+ (str.charAt(i) - ``'0'``);` `            ``rem = num % ``11``;` `        ``}` `     `  `        ``return` `rem;` `    ``}` `     `  `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``String str = ``"3435346456547566345436457867978"``;` `        ``System.out.println(remainder(str));` `    ``}` `}`   `/*This code is contributed by Nikita Tiwari.*/`

## Python3

 `# Python 3 implementation to find remainder` `# when a large number is divided by 11`   `# Function to return remainder` `def` `remainder(st) :` `    `  `    ``# len is variable to store the` `    ``# length of number string.` `    ``ln ``=` `len``(st)` `    `  `    ``rem ``=` `0` `    `  `    ``# loop that find remainder` `    ``for` `i ``in` `range``(``0``, ln) :` `        ``num ``=` `rem ``*` `10` `+` `(``int``)(st[i])` `        ``rem ``=` `num ``%` `11` `        `  `    ``return` `rem` `    `  `    `  `# Driver code` `st ``=` `"3435346456547566345436457867978"` `print``(remainder(st))`     `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# implementation to find remainder` `// when a large number is divided by 11` `using` `System;`   `class` `GFG` `{` `    `  `    ``// Function to return remainder` `    ``static` `int` `remainder(``string` `str)` `    ``{` `        `  `        ``// len is variable to store the` `        ``// length of number string.` `        ``int` `len = str.Length;` `    `  `        ``int` `num, rem = 0;` `    `  `        ``// loop that find remainder` `        ``for` `(``int` `i = 0; i < len; i++) ` `        ``{` `            ``num = rem * 10 + (str[i] - ``'0'``);` `            ``rem = num % 11;` `        ``}` `    `  `        ``return` `rem;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `str = ``"3435346456547566345436457867978"``;` `        ``Console.WriteLine(remainder(str));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`4`

Time Complexity: O(L ) where L is length of the string
Auxiliary Space: O(L )

Another Approach:

To find the remainder when a large number is divided by 11, we can use the following rule:

• Starting from the rightmost digit, we alternate between adding and subtracting the digits. That is, we add the rightmost digit, subtract the next digit to the left, add the next digit, and so on.
• If the result is divisible by 11, the original number is also divisible by 11, and the remainder is 0.
• Otherwise, the remainder is the absolute value of the result modulo 11.

## C++

 `#include` `using` `namespace` `std;`   `int` `remainder(string str)` `{` `    ``int` `n = str.length();` `    ``int` `sum = 0;` `    ``for` `(``int` `i = n-1; i >= 0; i--) {` `        ``int` `digit = str[i] - ``'0'``;` `        ``if` `((n-i) % 2 == 0) {` `            ``sum = (sum - digit + 11) % 11;` `        ``} ``else` `{` `            ``sum = (sum + digit) % 11;` `        ``}` `    ``}` `    ``return` `sum;` `}`   `int` `main()` `{` `    ``string str = ``"3435346456547566345436457867978"``;` `    ``cout << remainder(str);` `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main ` `{` `    ``public` `static` `int` `remainder(String str) ` `    ``{` `        ``int` `n = str.length();` `        ``int` `sum = ``0``;` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) ` `        ``{` `            ``int` `digit = str.charAt(i) - ``'0'``;` `            ``if` `((n - i) % ``2` `== ``0``) ` `            ``{` `                ``sum = (sum - digit + ``11``) % ``11``;` `            ``} ` `            ``else` `            ``{ ` `                ``sum = (sum + digit) % ``11``;` `            ``}` `        ``}` `        ``return` `sum;` `    ``}`   `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``String str = ``"3435346456547566345436457867978"``;` `        ``System.out.println(remainder(str));` `    ``}` `}`

## Python3

 `def` `remainder(``str``):` `    ``n ``=` `len``(``str``)` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(n``-``1``, ``-``1``, ``-``1``):` `        ``digit ``=` `int``(``str``[i])` `        ``if` `(n``-``i) ``%` `2` `=``=` `0``:` `            ``sum` `=` `(``sum` `-` `digit ``+` `11``) ``%` `11` `        ``else``:` `            ``sum` `=` `(``sum` `+` `digit) ``%` `11` `    ``return` `sum`   `str` `=` `"3435346456547566345436457867978"` `print``(remainder(``str``))`

Output

`4`

Time Complexity: O(n)
Auxiliary Space: O(1)

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