Look and Say - Google Top Interview Questions

                        Solution in C++ :

string solve(int n) {
    string s = "1", t;
    for (int i = 1; i < n; i++) {
        for (int j = 0, k = 0; j < s.size();) {
            char c = s[k];
            k++;
            while (k < s.size() && s[k] == s[k - 1]) k++;
            t.append(to_string(k - j));
            t.push_back(c);
            j = k;
        }
        t.swap(s);
        t.clear();
    }
    return s;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public String solve(int n) {
        // base case
        String s = "1";

        // for every k after 1, we need to apply the logic
        for (int k = 1; k < n; k++) {
            StringBuilder next = new StringBuilder();

            // count starts at 1, as we compare previous with current
            int count = 1;
            for (int i = 1; i <= s.length(); i++) {
                // if we reached the end of the string, or the char has changed,
                // we append the count and the previous char. restart the count to 1
                if (i == s.length() || s.charAt(i) != s.charAt(i - 1)) {
                    next.append(count).append(s.charAt(i - 1));
                    count = 1;
                } else {
                    // same char, just increase the count
                    count++;
                }
            }

            s = next.toString();
        }

        return s;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, n):
        s = "1"
        for _ in range(n - 1):
            temp = ""
            for x, y in groupby(s):
                temp += str(len(list(y))) + x
            s = temp
        return s