SAMSUNG 49-Inch Odyssey G9

Because earth is not flat

Description

The "Count and Say" sequence is a sequence of integers where each term is derived from describing the digits of the previous term. Starting with "1", the sequence progresses by reading off the digits of the previous term, counting the number of digits in groups, and then describing them.

For example:

• 1st term: "1"
• 2nd term: "11" (one 1)
• 3rd term: "21" (two 1s)
• 4th term: "1211" (one 2, then one 1)
• 5th term: "111221" (one 1, one 2, then two 1s)

The goal is to generate the nth term in this sequence.

Intuition

The key insight is to observe how each term in the sequence describes the previous term. Each term is formed by counting consecutive digits in the previous term and then stating the count followed by the digit. The approach is iterative, where the output of one step becomes the input for the next.

The challenge lies in properly managing the counting and ensuring that the sequence is correctly described for each step, especially as the terms get longer.

Approach

1. Initialization:

   • Start with the first term "1".

2. Iterative Generation:

   • For each subsequent term, iterate over the digits of the current term.
   • Track the current digit and count how many times it repeats consecutively.
   • Once a different digit is encountered, append the count and the digit to form part of the next term.
   • Reset the counter and update the current digit.
   • After iterating through all digits, append the final count and digit to complete the next term.

3. Edge Cases:

   • Handle cases where the term is very short (e.g., "1") or where the entire term is made of a single repeating digit.

4. Output:

   • After generating terms iteratively up to the nth term, return the final term.

Complexity

Time Complexity:

The time complexity is difficult to express precisely in Big-O notation because it depends on how the sequence evolves. However, each term roughly doubles in length with each iteration, leading to a time complexity of O(2^n) for the nth term. Space Complexity:

space complexity:

O(m) where m is the length of the term being generated, as we need to store the string and the intermediate counts.

Code

C++

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

        for (int i = 0; i < n - 1; i++) {
            string str = "";
            char prevChar = out[0];
            int count = 1;

            for (int j = 1; j < out.length(); j++) {
                if (out[j] == prevChar) {
                    count++;
                } else {
                    str += to_string(count) + prevChar;
                    prevChar = out[j];
                    count = 1;
                }
            }

            str += to_string(count) + prevChar;
            out = str;
        }

        return out;
    }
};

Python

class Solution:
    def countAndSay(self, n: int) -> str:
        out = "1"

        for _ in range(n - 1):
            str_out = ""
            prev_char = out[0]
            count = 1

            for j in range(1, len(out)):
                if out[j] == prev_char:
                    count += 1
                else:
                    str_out += str(count) + prev_char
                    prev_char = out[j]
                    count = 1

            str_out += str(count) + prev_char
            out = str_out

        return out

Java

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

        for (int i = 0; i < n - 1; i++) {
            StringBuilder str = new StringBuilder();
            char prevChar = out.charAt(0);
            int count = 1;

            for (int j = 1; j < out.length(); j++) {
                if (out.charAt(j) == prevChar) {
                    count++;
                } else {
                    str.append(count).append(prevChar);
                    prevChar = out.charAt(j);
                    count = 1;
                }
            }

            str.append(count).append(prevChar);
            out = str.toString();
        }

        return out;
    }
}

JavaScript

var countAndSay = function (n) {
    var out = "1";

    for (let i = 0; i < n - 1; i++) {
        var prevChar = out.charAt(0);
        var count = 1;
        var str = "";
        for (let j = 1; j < out.length; j++) {
            if (prevChar == out.charAt(j)) {
                count++;
            }
            else {
                str += count + "" + prevChar;
                prevChar = out.charAt(j);
                count = 1;
            }
        }
        str += count + "" + prevChar;
        out = str;
    }
    return out;
};