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Count ways to spell a number with repeated digits
• Difficulty Level : Easy
• Last Updated : 31 Mar, 2021

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++

 // C++ program to count number of ways we// can spell a number#includeusing namespace std;typedef long long int ll; // Function to calculate all possible spells of// a number with repeated digits// num --> string which is favourite numberll 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

## Java

 // Java program to count number of ways we// can spell a numberclass 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.

## Python3

 # 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 numberdef spellsCount(num):     n = len(num);     # final count of total    # possible spells    result = 1;     # iterate through complete    # number    i = 0;    while(i

## C#

 // C# program to count number of ways we// can spell a numberusing 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.

## PHP

 string// which is favourite numberfunction 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.?>

## Javascript



Output :

8

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