For a given n, Print the count and say sequence using c++

Pre-requisite:

to_string function: In C++ we have an in-build function which converts a numerical value to its string representation. Like 1 is converted to "1" (the first 1 is of int datatype, while the second is of string datatype).

prototype of the function is:

    string to_string (int value);

Algorithm:

1. Check for base cases

    if(n==0)
        return "";
    if(n==1)
        return "1";

1. For rest of the cases, it must start from 1, thus initialize result with string "1". result is our output string which will be printed for each label.
2. Print result for level 1. (As level >1 now)
3. Declare a temporary string only to get the current level result. Let the temporary string be temp.
4. For i=1:n //iterate for rest of the rows
   1. Store the length of result (carrying the result of last processed level actually) string length
   2. Let the length be len
   3. For j=0:len
      Initialize a count variable to store repeating number's frequency
      While same number( character in result string actually) is repeating
      Increment count
      End While
      Convert count value to string using the to_string function.
      temp=temp+ to_string(count) + result[j] //count comes first
      End FOR
   4. Current level string is processed in temp, So update result=temp
   5. Print result
   6. Clear temp for further levels
5. End For

C++ implementation

#include <bits/stdc++.h>
using namespace std;

void countAndSay(int n) {
	//base cases
	if(n==0)
		return "";
	if(n==1)
		return "1";
	//for rest of the cases, it must start from 1   
	//initialize result with string "1"
	string result="1";
	cout<<result<<endl;
	string temp;//temporary string to hold levelwise result
	for(int i=1;i<n;i++){ //iterate for n-1 rows
		//iterate upto current string length
		int len=result.length();
		for(int j=0;j<len;j++){
			int count=1;//initialize count as 1
			//find count for repeated number
			while(j+1<len && result[j]==result[j+1] ){
				count++;
				j++;
			}
			//convert the count to string and add to 
			//temprary result, then add original no
			temp+=to_string(count)+result[j];
		}
		//assign temporary result to original result 
		//& print for current level 
		result=temp;
		cout<<result<<endl;
		//clear the temporary result
		temp="";
	}
}

int main(){
	int n;
	
	cout<<"count and Say problem.....\n";
	cout<<"enter n, no of rows\n";
	cin>>n;
	//function to print count and say sequence
	coutAndSay(n);
	
	return 0;
}

Output

First run:
count and Say problem.....
enter n, no of rows
3
Printing Count and Say sequence...
1
11
21


Second run:
enter n, no of rows
6
Printing Count and Say sequence...
1
11
21
1211
111221
312211

Third run:
count and Say problem.....
enter n, no of rows
10
Printing Count and Say sequence...
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211