Count and Say | Leetcode 38

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

        string number = "1";
        for(int i=2;i<=n;++i){
            //Build new string
            string res;
            int count = 1;
            char curr = number[0];
            for(int j=1;j<number.size();++j){
                if(number[j]==curr) count++;
                else{
                    res += to_string(count) + curr;
                    curr = number[j];
                    count = 1;
                }
            }
            res += to_string(count) + curr;
            number = res;
        }
        return number;
    }
};

/*
//JAVA
class Solution {
    public String countAndSay(int n) {
        if (n == 1) return "1";
        
        StringBuilder number = new StringBuilder("1");
        for (int i = 2; i <= n; i++) {
            StringBuilder res = new StringBuilder();
            int count = 1;
            char curr = number.charAt(0);
            for (int j = 1; j < number.length(); j++) {
                if (number.charAt(j) == curr) {
                    count++;
                } else {
                    res.append(count).append(curr);
                    curr = number.charAt(j);
                    count = 1;
                }
            }
            res.append(count).append(curr);
            number = res;
        }
        return number.toString();
    }
}

#Python
class Solution:
    def countAndSay(self, n: int) -> str:
        if n == 1:
            return "1"
        
        number = "1"
        for _ in range(2, n + 1):
            res = []
            count = 1
            curr = number[0]
            for j in range(1, len(number)):
                if number[j] == curr:
                    count += 1
                else:
                    res.append(str(count) + curr)
                    curr = number[j]
                    count = 1
            res.append(str(count) + curr)
            number = "".join(res)
        return number
*/