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++
// 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; } |
chevron_right
filter_none
Java
// 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 |
chevron_right
filter_none
Python
# 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 |
chevron_right
filter_none
C#
// 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. |
chevron_right
filter_none
Output:
1321123113
GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details
Recommended Posts:
- Convert a number of length N such that it contains any one digit at least 'K' times
- Minimum steps to convert one binary string to other only using negation
- Generate a number such that the frequency of each digit is digit times the frequency in given number
- Convert time from 24 hour clock to 12 hour clock format
- Find minimum number of steps to reach the end of String
- Find the Number which contain the digit d
- Sum of digit of a number using recursion
- Program to Convert BCD number into Decimal number
- Find the frequency of a digit in a number
- Program to delete Nth digit of a Number
- Perform the given queries on the rooted tree
- C++ Program for Smallest K digit number divisible by X
- C++ Program for Largest K digit number divisible by X
- Find the minimum number possible by changing at most one digit
- Count the occurrence of digit K in a given number N using Recursion
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
