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Leetcode 38. Count and Say

Problem Statement

The “Count and Say” sequence is a sequence of digit strings defined by the following recursive formula:

• countAndSay(1) is "1".
• To generate the next term in the sequence from the previous term, count the number of consecutive digits and say the digit. For example, the term after "1" is "11", because we have one 1 (“one 1” hence “11”).

Given an integer n, generate the n-th term of the “Count and Say” sequence.

Clarifying Questions

1. What is the maximum value of n we need to consider?
   • This will guide the performance requirements.
2. Is there any specific form to return the value such as a string or an integer?
   • Typically this would be a string.
3. Are there any constraints on generating intermediate results regarding memory usage?

Strategy

1. Base Case: Define the base case for n = 1, which is “1”.
2. Recursive/Iterative Approach: Build the sequence iteratively from 1 to n.
   • Use a loop to process the previous term character by character.
   • Count consecutive characters and append the count followed by the respective character to build the next term.
3. Edge Cases: Consider cases like n = 0 which should return an empty string, although the problem statement starts counting from 1.

Here is the implementation to solve the problem.

Code

#include <iostream>
#include <string>

class Solution {
public:
    std::string countAndSay(int n) {
        if (n == 1) return "1";
        
        std::string result = "1";
        for (int i = 2; i <= n; ++i) {
            std::string current = "";
            int len = result.size();
            for (int j = 0; j < len; ++j) {
                int count = 1;
                while (j + 1 < len && result[j] == result[j + 1]) {
                    ++count;
                    ++j;
                }
                current += std::to_string(count) + result[j];
            }
            result = current;
        }
        return result;
    }
};

// Example usage:
// int main() {
//     Solution sol;
//     int n = 4;
//     std::cout << sol.countAndSay(n) << std::endl;  // Output should be "1211"
//     return 0;
// }

Time Complexity

• Time Complexity: O(m * n), where n is the given input and m is the maximum length of any string in the sequence up to n. In the worst case, the length of the string could approximately double at each step (though it’s typically less than that in practice).
• Space Complexity: O(m), where m is the length of the string at each step since we only build the next string based on the current one.

This approach ensures that the sequence is generated efficiently and handles varying sizes of n smoothly.

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