Count ways to spell a number with repeated digits

Given a string that contains digits of a number. The number may contain many same continuous digits in it. The task is to count number of ways to spell the number.
For example, consider 8884441100, one can spell it simply as triple eight triple four double two and double zero. One can also spell as double eight, eight, four, double four, two, two, double zero.

Examples :

Input :  num = 100
Output : 2
The number 100 has only 2 possibilities, 
1) one zero zero 
2) one double zero.

Input : num = 11112
Output: 8
1 1 1 1 2, 11 1 1 2, 1 1 11 2, 1 11 1 2
11 11 2, 1 111 2, 111 1 2, 1111 2

Input : num = 8884441100
Output: 64

Input : num = 12345
Output: 1

Input : num = 11111
Output: 16

This is a simple problem of permutation and combination. If we take example test case given in the question, 11112. The answer depends on the number of possible combinations of 1111. The number of combinations of “1111” is 2^3 = 8. As our combinations will depend on whether we choose a particular 1 and for “2” there will be only one possibility 2^0 = 1, so answer for “11112” will be 8*1 = 8.
So, the approach is to count the particular continuous digit in string and multiply 2^(count-1) with previous result.

C++

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// C++ program to count number of ways we
// can spell a number
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
  
// Function to calculate all possible spells of
// a number with repeated digits
// num --> string which is favourite number
ll spellsCount(string num)
{
    int n = num.length();
  
    // final count of total possible spells
    ll result = 1;
  
    // iterate through complete number
    for (int i=0; i<n; i++)
    {
       // count contiguous frequency of particular
       // digit num[i]
       int count = 1;
       while (i < n-1 && num[i+1] == num[i])
       {
           count++;
           i++;
       }
  
       // Compute 2^(count-1) and multiply with result  
       result = result * pow(2, count-1);
    }
    return result;
}
  
// Driver program to run the case
int main()
{
    string num = "11112";
    cout << spellsCount(num);
    return 0;
}

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Java

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// Java program to count number of ways we
// can spell a number
class GFG {
      
    // Function to calculate all possible 
    // spells of a number with repeated digits
    // num --> string which is favourite number
    static long spellsCount(String num)
    {
          
        int n = num.length();
  
        // final count of total possible spells
        long result = 1;
  
        // iterate through complete number
        for (int i = 0; i < n; i++) {
              
            // count contiguous frequency of 
            // particular digit num[i]
            int count = 1;
              
            while (i < n - 1 && num.charAt(i + 1
                               == num.charAt(i)) {
                                      
                count++;
                i++;
            }
  
            // Compute 2^(count-1) and multiply 
            // with result
            result = result * 
                     (long)Math.pow(2, count - 1);
        }
        return result;
    }
  
    public static void main(String[] args)
    {
  
        String num = "11112";
  
        System.out.print(spellsCount(num));
    }
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python3 program to count number of
# ways we can spell a number
  
# Function to calculate all possible 
# spells of a number with repeated 
# digits num --> string which is 
# favourite number
def spellsCount(num):
  
    n = len(num);
  
    # final count of total
    # possible spells
    result = 1;
  
    # iterate through complete
    # number
    i = 0;
    while(i<n):
      
        # count contiguous frequency 
        # of particular digit num[i]
        count = 1;
        while (i < n - 1 and 
               num[i + 1] == num[i]):
      
            count += 1;
            i += 1;
  
        # Compute 2^(count-1) and
        # multiply with result 
        result = result * int(pow(2, count - 1));
        i += 1;
    return result;
  
# Driver Code
num = "11112";
print(spellsCount(num));
  
# This code is contributed
# by mits

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C#

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// C# program to count number of ways we
// can spell a number
using System;
  
class GFG {
      
    // Function to calculate all possible 
    // spells of a number with repeated 
    // digits num --> string which is
    // favourite number
    static long spellsCount(String num)
    {
          
        int n = num.Length;
  
        // final count of total possible
        // spells
        long result = 1;
  
        // iterate through complete number
        for (int i = 0; i < n; i++)
        {
              
            // count contiguous frequency of 
            // particular digit num[i]
            int count = 1;
              
            while (i < n - 1 && num[i + 1] 
                                == num[i])
            {
                count++;
                i++;
            }
  
            // Compute 2^(count-1) and multiply 
            // with result
            result = result * 
                    (long)Math.Pow(2, count - 1);
        }
          
        return result;
    }
  
    // Driver code
    public static void Main()
    {
  
        String num = "11112";
  
        Console.Write(spellsCount(num));
    }
}
  
// This code is contributed by nitin mittal.

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PHP

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<?php
// PHP program to count 
// number of ways we
// can spell a number
  
// Function to calculate 
// all possible spells of
// a number with repeated 
// digits num --> string
// which is favourite number
function spellsCount($num)
{
    $n = strlen($num);
  
    // final count of total
    // possible spells
    $result = 1;
  
    // iterate through 
    // complete number
    for ($i = 0; $i < $n; $i++)
    {
    // count contiguous frequency 
    // of particular digit num[i]
    $count = 1;
    while ($i < $n - 1 && 
           $num[$i + 1] == $num[$i])
    {
        $count++;
        $i++;
    }
  
    // Compute 2^(count-1) and
    // multiply with result 
    $result = $result
              pow(2, $count - 1);
    }
    return $result;
}
  
// Driver Code
$num = "11112";
echo spellsCount($num);
  
// This code is contributed
// by nitin mittal. 
?>

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Output :

8

If you have another approach to solve this problem then please share.

This article is contributed by Shashank Mishra ( Gullu ). 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.

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Improved By : nitin mittal, Mithun Kumar