Perform n steps to convert every digit of a number in the format [count][digit]

Given a number num as a string and a number N. The task is to write a program which converts the given number num to another number after performing N steps. At each step, every digit of num will be written in the format [count][digit] in the new number, where count is the number of times a digit occurs consecutively in num.

Examples:

Input: num = “123”; n = 3
Output: 1321123113
For, n = 1: 123 becomes 1 time 1, 1 time 2, 1 time 3, hence number 111213
For, n = 2: 3 times 1, 1 time 2, 1 time 1, 1 time 3, hence number 31121113
For, n = 3: 1 time 3, 2 times 1, 1 time 2, 3 times 1, 1 time 3, hence number 1321123113

Input: num = “1213”; n = 1
Output: 11121113

Approach: Parse the string’s characters as a single digit and maintain a count for that digit till a different digit is found. Once a different digit is found, add the count of the digit to the new string and number to it. Once the string is parsed completely, recur for the function again with this new string till n steps are done.



Below is the implementation of the above approach:

C++

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// C++ program to convert number
// to the format [count][digit] at every step
#include <bits/stdc++.h>
using namespace std;
  
// Function to perform every step
void countDigits(string st, int n)
{
  
    // perform N steps
    if (n > 0) {
        int cnt = 1, i;
        string st2 = "";
  
        // Traverse in the string
        for (i = 1; i < st.length(); i++) {
            if (st[i] == st[i - 1])
                cnt++;
            else {
                st2 += ('0' + cnt);
                st2 += st[i - 1];
                cnt = 1;
            }
        }
  
        // for last digit
        st2 += ('0' + cnt);
        st2 += st[i - 1];
  
        // recur for current string
        countDigits(st2, --n);
    }
  
    else
        cout << st;
}
  
// Driver Code
int main()
{
  
    string num = "123";
    int n = 3;
  
    countDigits(num, n);
  
    return 0;
}

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Java

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// Java program to convert number 
// to the format [count][digit] at every step
class GFG 
{
  
    // Function to perform every step
    public static void countDigits(String st, int n)
    {
  
        // perform N steps
        if (n > 0
        {
            int cnt = 1, i;
            String st2 = "";
  
            // Traverse in the string
            for (i = 1; i < st.length(); i++) 
            {
                if (st.charAt(i) == st.charAt(i - 1))
                    cnt++;
                else
                {
                    st2 += ((char) 0 + (char) cnt);
                    st2 += st.charAt(i - 1);
                    cnt = 1;
                }
            }
  
            // for last digit
            st2 += ((char) 0 + (char) cnt);
            st2 += st.charAt(i - 1);
  
            // recur for current string
            countDigits(st2, --n);
        
        else
            System.out.print(st);
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        String num = "123";
        int n = 3;
        countDigits(num, n);
    }
}
  
// This code is contributed by
// sanjeev2552

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Python

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# Python program to convert number
# to the format [count][digit] at every step
  
# Function to perform every step
def countDigits(st, n):
  
    # perform N steps
    if (n > 0) :
        cnt = 1
        i = 0
        st2 = ""
        i = 1
          
        # Traverse in the string
        while (i < len(st) ) :
            if (st[i] == st[i - 1]):
                cnt = cnt + 1
            else :
                st2 += chr(48 + cnt)
                st2 += st[i - 1]
                cnt = 1
            i = i + 1
  
        # for last digit
        st2 += chr(48 + cnt)
        st2 += st[i - 1]
  
        # recur for current string
        countDigits(st2, n - 1)
        n = n - 1;
  
    else:
        print(st)
  
# Driver Code
  
num = "123"
n = 3
  
countDigits(num, n)
  
# This code is contributed by Arnab Kundu

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C#

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// C# program to convert number 
// to the format [count][digit] at every step
using System;
class GFG 
{
  
// Function to perform every step
public static void countDigits(string st, int n)
{
  
    // perform N steps
    if (n > 0) 
    {
        int cnt = 1, i;
        string st2 = "";
  
        // Traverse in the string
        for (i = 1; i < st.Length; i++) 
        {
            if (st[(i)] == st[(i - 1)])
                cnt++;
            else
            {
                st2 += ((char) 0 + (char) cnt);
                st2 += st[(i - 1)];
                cnt = 1;
            }
        }
  
        // for last digit
        st2 += ((char) 0 + (char) cnt);
        st2 += st[(i - 1)];
  
        // recur for current string
        countDigits(st2, --n);
    
    else
        Console.Write(st);
}
  
// Driver Code
public static void Main() 
{
    string num = "123";
    int n = 3;
    countDigits(num, n);
}
}
  
// This code is contributed by
// Code_Mech.

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Output:

1321123113

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