# Remove recurring digits in a given number

• Difficulty Level : Easy
• Last Updated : 01 Sep, 2022

Given a number as string, remove recurring digits from the given string. The changes must be made in-place. Expected time complexity O(n) and auxiliary space O(1).
Examples:

```Input:  num[] = "1299888833"
Output: num[] = "12983"

Input:  num[] = "1299888833222"
Output: num[] = "129832"```

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This problem is similar to Run Length Encoding

```Let num[] be input number represented as character array

1) Initialize index of modified string 'j' as 0.
2) Traverse input string and do following for every digit num[i].
a) Copy current character 'num[i]' to 'num[j]' and increment i & j.
b) Keep incrementing i while num[i] is same as previous digit.
3) Add string termination character at 'num[j]'```

Below is the implementation of above algorithm.

## C++

 `// C++ program to remove recurring digits from``// a given number``#include ``using` `namespace` `std;` `/* Removes recurring digits in num[]  */``void` `removeRecurringDigits(``char` `num[])``{``    ``int` `len = ``strlen``(num);` `    ``int` `j = 0; ``// Index in modified string` `    ``/* Traverse digits of given number one by one */``    ``for` `(``int` `i=0; i

## Java

 `// Java program to remove recurring``// digits from a given number``class` `GFG``{` `    ``/* Removes recurring digits in num[] */``    ``static` `String removeRecurringDigits(``char` `num[])``    ``{``        ``int` `len = num.length;` `        ``int` `j = ``0``; ``// Index in modified string``        ``String s = ``""``;``        ` `        ``/* Traverse digits of given number one by one */``        ``for` `(``int` `i = ``0``; i < len; i++)``        ``{``            ` `            ``/* Copy the first occurrence of new digit */``            ``s += String.valueOf(num[i]);` `            ``/* Remove repeating occurrences of digit */``            ``while` `(i + ``1` `< len && num[i] == num[i + ``1``])``            ``{``                ``i++;``            ``}``        ``}``        ``return` `s;``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``char` `num[] = ``"1299888833"``.toCharArray();``        ``System.out.print(``"Modified number is "` `+``                        ``removeRecurringDigits(num));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python3 program to remove recurring``# digits from a given number` `# Removes recurring digits in num[]``def` `removeRecurringDigits(num):` `    ``l ``=` `len``(num)``    ` `    ``# Index in modified string``    ``(i, j) ``=` `(``0``, ``0``)``    ``str` `=` `''``    ` `    ``# Traverse digits of given``    ``# number one by one``    ``while` `i < l:``        ` `        ``# Copy the first occurrence``        ``# of new digit``        ``str` `+``=` `num[i]``        ``j ``+``=` `1``        ` `        ``# Remove repeating occurrences of digit``        ``while` `(i ``+` `1` `< l ``and` `num[i] ``=``=` `num[i ``+` `1``]):``            ``i ``+``=` `1``        ``i ``+``=` `1` `    ``return` `str` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``num ``=` `'1299888833'``    ``print``(``'Modified number is {}'``.``format``(``           ``removeRecurringDigits(num)))` `# This code is contributed by rutvik_56`

## C#

 `// C# program to remove recurring``// digits from a given number``using` `System;` `class` `GFG``{` `    ``/* Removes recurring digits in num[] */``    ``static` `String removeRecurringDigits(``char` `[]num)``    ``{``        ``int` `len = num.Length;` `        ``int` `j = 0; ``// Index in modified string``        ``String s = ``""``;``        ` `        ``/* Traverse digits of given number one by one */``        ``for` `(``int` `i = 0; i < len; i++)``        ``{``            ` `            ``/* Copy the first occurrence of new digit */``            ``s += String.Join(``""``,num[i]);` `            ``/* Remove repeating occurrences of digit */``            ``while` `(i + 1 < len && num[i] == num[i + 1])``            ``{``                ``i++;``            ``}``        ``}``        ``return` `s;``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `Main()``    ``{``        ``char` `[]num = ``"1299888833"``.ToCharArray();``        ``Console.Write(``"Modified number is "` `+``                        ``removeRecurringDigits(num));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`Modified number is 12983`

Time complexity: O(N2) where N is the length of the string.

Auxiliary space complexity: O(1).

Method 2:

The main idea behind this approach is that we will traverse the string check next character of all index from 0 to n-2.

If the value at both indexes is the same then we will delete anyone of them. so, the length of the string will get reduced by 1.

Below is the implementation of the above approach.

## C++

 `//C++ program to remove reccuring``//degits in a given number``#include``using` `namespace` `std;``void` `updatestring(string s){``    ``int` `x=s.length()-2;``    ``//traversing the string``    ``for``(``int` `i=0;is.length()-2){``            ``break``;``        ``}``        ``if``(s[i]==s[i+1]){``            ``//removing character which is``            ``//occurring more then one time``            ``s.erase(s.begin()+i);``            ``i--;``        ``}``    ``}``    ``cout<<``"number without reccuring degits: "``<

## Python3

 `# Python3 program to remove recurring``# digits from a given number` `# Removes recurring digits in num[]``def` `removeRecurringDigits(s):``    ``x ``=` `len``(s) ``-` `2``    ``# traversing the string``    ``for` `i ``in` `range``(x):``        ``if``(i > ``len``(s) ``-` `2``):``            ``break``        ``if``(s[i]``=``=``s[i``+``1``]):``            ``# removing character which is``            ``# occurring more then one time``            ``s ``=` `s[:i] ``+`  `s[i``+``1``:]``            ``i ``=` `i ``-` `1``    ``return` `s` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``num ``=` `'1299888833'``    ``print``(``'Modified number is {}'``.``format``(``        ``removeRecurringDigits(num)))``    ` `    ``# This code is contributed by 111arpit1.`

Output

`number without reccuring degits: 129832`

Time complexity: O(N) where N is the length of the string.

Auxiliary space complexity: O(1).

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