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Remove recurring digits in a given number

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  • Difficulty Level : Easy
  • Last Updated : 01 Sep, 2022
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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 <bits/stdc++.h>
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<len; i++)
    {
        /* Copy the first occurrence of new digit */
        num[j++] = num[i];
 
        /* Remove repeating occurrences of digit */
        while (i + 1 < len && num[i] == num[i+1])
            i++;
    }
 
    /* terminate the modified string */
   num[j] = '\0';
}
 
/* Driver program to test above function */
int main()
{
    char num[] = "1299888833";
    removeRecurringDigits(num);
    cout << "Modified number is " << num;
    return 0;
}

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




<script>
// Javascript program to remove recurring
// digits from a given number
 
 /* Removes recurring digits in num[] */
function removeRecurringDigits(num)
{
    let len = num.length;
   
        let j = 0; // Index in modified string
        let s = "";
           
        /* Traverse digits of given number one by one */
        for (let i = 0; i < len; i++)
        {
               
            /* Copy the first occurrence of new digit */
            s += (num[i]);
   
            /* Remove repeating occurrences of digit */
            while (i + 1 < len && num[i] == num[i + 1])
            {
                i++;
            }
        }
        return s;
}
 
/* Driver code */
let num="1299888833".split("");
document.write("Modified number is " +
                        removeRecurringDigits(num));
 
 
// This code is contributed by rag2127
</script>

Output

Modified number is 12983

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

Auxiliary space complexity: O(1). 
 

This article is contributed by Priyanka. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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<bits/stdc++.h>
using namespace std;
void updatestring(string s){
    int x=s.length()-2;
    //traversing the string
    for(int i=0;i<x;i++){
         
        if(i>s.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: "<<s<<endl;
}
//driver code
int main() {
string s="1299888833222";
//function call
updatestring(s);
return 0;
}
 
//this code is contribut by Machhaliya Muhammad

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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