Count and Say


Problem Statement :


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 substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

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


Solution :



title-img 


                            Solution in C :

char* countAndSay(int n)
{
    if(n == 1)
    {
        return "1";
    }
    // get s
    char* s = countAndSay(n - 1);  // answer from previous call
    int size = strlen(s);  // size of s
    // get ans
    char* ans = (char*) malloc(4463 * sizeof(char));
    // iterate through s and add to ans
    int idx = 0;  // idx in ans
    char curr = *s;  // first char of s
    int cnt = 1;  // count of curr
    for(int i = 1; i < size; i++)
    {
        char c = *(s + i);
        if(c == curr)
        {
            cnt++;
        }
        else
        {
            *(ans + idx++) = ('0' + cnt);
            *(ans + idx++) = curr;
            curr = c;
            cnt = 1;
        }
        // printf("%s\n", ans);
    }
    *(ans + idx++) = ('0' + cnt);
    *(ans + idx++) = curr;
    *(ans + idx) = '\0';
    // ans = (char*) realloc(ans, (idx + 1) * sizeof(char));
    return ans;
}
                        


                        Solution in C++ :

class Solution {
public:
    string countAndSay(int n) {
        if(n == 1)  return "1";

        string s = countAndSay(n-1);

        int i = 0, j = 0;
        string r;

        while(i < s.size()) {
            j = i+1;

            while(j < s.size() && s[j] == s[i])
                ++j;

            string ct = to_string(j-i);
            r += ct + s[i];
            
            i = j;
        }
        return r;
    }
};
                    


                        Solution in Java :

class Solution {
    public String countAndSay(int n) {
        if(n==1)
        return "1";

        String s=countAndSay(n-1);

        int c=0;
        StringBuilder ans=new StringBuilder();

        for(int i=0;i<s.length();i++){
            c++;
            if(i==s.length()-1 || s.charAt(i)!=s.charAt(i+1)){
                ans.append(c).append(s.charAt(i));
                c=0;
            }
        }
        return ans.toString();
        
    }
}
                    


                        Solution in Python : 
                            
res = '1'
    if n == 1:
        return res
    
    for i in range(2,n+1):
        ans = []
        charPointer = res[0]
        countPointer = 0
        for char in res:
            if char == charPointer:
                countPointer += 1
            else:
                ans.append(str(countPointer))
                ans.append(charPointer)
                charPointer = char
                countPointer = 1
        ans.append(str(countPointer))
        ans.append(charPointer)
        res = "".join(ans)
    return res


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