Last updated on October 10th, 2024 at 12:04 am

Here, We see Count and Say LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Count and Say LeetCode Solution

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"

1. Count and Say Leetcode Solution C++

class Solution {
public:
    string countAndSay(int n) {
        if(n == 1)
          return "1";
        if(n == 2)
          return "11";
          
        string str = "11";
        
        for(int i = 3; i <= n; i++)
         {
             str += '$';
             int len = str.length();
             
             int cnt = 1;
             string tmp = "";
             
             for(int j = 1; j < len; j++)
              {
                  if(str[j] != str[j - 1])
                    {
                        tmp += cnt + '0';
                        tmp += str[j - 1];
                        cnt = 1;
                    }
                   else
                     cnt++;
              }
            str = tmp;  
         }
         
        return str;       
    }
};

2. Count and Say Leetcode Solution Java

class Solution {
    public String countAndSay(int n) {
        String s = "1";
        for(int i = 1; i < n; i++){
            s = countIdx(s);
        }
        return s;
    }
    
    public String countIdx(String s){
        StringBuilder sb = new StringBuilder();
        char c = s.charAt(0);
        int count = 1;
        for(int i = 1; i < s.length(); i++){
            if(s.charAt(i) == c){
                count++;
            }
            else
            {
                sb.append(count);
                sb.append(c);
                c = s.charAt(i);
                count = 1;
            }
        }
        sb.append(count);
        sb.append(c);
        return sb.toString();        
    }
}

3. Count and Say Leetcode Solution JavaScript

var countAndSay = function(n) {
    var str = '1';
    for (var i=1; i < n; i++) {
        var strArray = str.split('');
        str ='';
        var count = 1;
        // Loop through current nth level line
        for (var j=0; j < strArray.length; j++) {
            // Next digit is different
            if (strArray[j] !== strArray[j+1]) {
                // Go to next non-matching digit
                str += count + strArray[j];
                count = 1;
            }
            else {
                count++;
            }
        }
    }
    return str;    
};

4. Count and Say Leetcode Solution Python

class Solution(object):
    def countAndSay(self, n):
        s = '1'
        for _ in range(n - 1):
            s = re.sub(r'(.)\1*', lambda m: str(len(m.group(0))) + m.group(1), s)
        return s