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We are given a number containing many repeated digits as a string. The goal is to find the number of ways to spell it. For example 112233 can be spelled as double one, double two double three or one one two two three three.

We will do this by checking continuous numbers. If the number is “13” there is only one way to spell it as “one three” (20). If digits are “113” ways “double one three”, “one one three” (21). So, the approach is to count the one continuous digit in string and multiply 2(count-1) with the previous result.

Let’s understand with examples.

**Input**

num=”11211”

**Output**

Count of ways to spell a number with repeated digits are: 4

**Explanation**

ways are: 1. One one two one one 2. Double one two one one 3. One one two double one 4. Double one two double one

**Input**

num=”2212”

**Output**

Count of ways to spell a number with repeated digits are: 2

**Explanation**

ways are: 1. Two two one two 2. Double two one two

We are taking string str to represent a number.

Function word_spell(string str) takes the number in str and returns ways to spell it.

Take the initial variable count as 0 for such ways

Using for loop traverse str for each digit.

Take a variable temp as the number of repetitions of a particular digit. If str[i]==str[i+1], increase temp.

Calculate count=count*pow(2,temp-1)

At the end return count as result.

#include<bits/stdc++.h> using namespace std; long long int word_spell(string str){ long long int count = 1; int len = str.length(); for (int i=0; i<len; i++){ int temp = 1; while(i < len-1 && str[i+1] == str[i]){ temp++; i++; } count = count * pow(2, temp-1); } return count; } int main(){ string str = "222211"; cout<<"Count of ways to spell a number with repeated digits are: "<<word_spell(str); return 0; }

If we run the above code it will generate the following output −

Count of ways to spell a number with repeated digits are: 16

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