Leetcode 38 Count and say solution

38. Count and Say

Medium

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• countAndSay(1) = "1"
• countAndSay(n) is the way you would "say" the digit string from countAndSay(n-1), which is then converted into a different digit string.

To determine how you "say" a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

For example, the saying and conversion for digit string "3322251":

Given a positive integer n, return the nth term of the count-and-say sequence.

 

Example 1:

Input: n = 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"

 

Constraints:

• 1 <= n <= 30

Practice Here :

(9) Count and Say - LeetCode

Solution :

#include <bits/stdc++.h>

class Solution {

public:

    string say(string n)

    {   string ans="";

        int ct =1;

        char c = n[0];

        for(int i=1;i<n.length();i++)

        {

            if(n[i]!=c)

            {

               ans+=(ct+'0');

                ans+=(c);

                ct =1;

                c = n[i];

            }else {

                ct++;

            }

        }

     ans+=(ct+'0');

        ans+=(c);

     return ans;

    }

    string countAndSay(int n) {

       vector<string> v;

        v.push_back("0");

        v.push_back("1");

        for(int i=2;i<=n;i++)

        {  

            v.push_back(say(v[i-1]));

        }

        return v[n];

    }

};