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 1321123113Input: 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; } |
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 |
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 |
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. |
1321123113
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