Count and Say | Made Super Easy | Simple Explanation | Leetcode 38 | codestorywithMIK
Count and Say - Video Solution
Watch 10 video solutions for Count and Say, a medium level problem involving String. This walkthrough by codestorywithMIK has 51,962 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text 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 run-length encoding ofcountAndSay(n - 1).
Run-length encoding (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "3322251" we replace "33" with "23", replace "222" with "32", replace "5" with "15" and replace "1" with "11". Thus the compressed string becomes "23321511".
Given a positive integer n, return the nth element of the count-and-say sequence.
Example 1:
Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1" countAndSay(2) = RLE of "1" = "11" countAndSay(3) = RLE of "11" = "21" countAndSay(4) = RLE of "21" = "1211"
Example 2:
Input: n = 1
Output: "1"
Explanation:
This is the base case.
Constraints:
1 <= n <= 30
Follow up: Could you solve it iteratively?
Approach Overview
Problem Overview: Generate the nth term of the Count and Say sequence. Each term is created by reading the previous term and describing consecutive digits ("one 1", "two 1s", etc.), producing the next string.
Approach 1: Iterative String Simulation (O(n * L) time, O(L) space)
Build the sequence step by step starting from "1". For every iteration, scan the current string and group consecutive identical digits. Count how many times a digit repeats, append count + digit to a new string, then continue scanning. The key operation is a linear pass with two pointers (or an index and counter) to detect runs of the same character. If the length of the nth term is L, the total work is O(n * L) time with O(L) space for the constructed string.
This method is essentially string simulation. No advanced data structures are required—just careful iteration and string building. Problems involving sequence generation like this often fall under string manipulation and simulation. Iterative construction avoids recursion overhead and is straightforward to debug.
Approach 2: Recursive Construction (O(n * L) time, O(n + L) space)
Define the problem recursively: the nth term depends entirely on the (n-1)th term. First compute countAndSay(n-1), then run the same grouping logic over that returned string to build the nth result. The recursion depth is n, and each level processes a string of length L, so the time complexity remains O(n * L). Additional space comes from the call stack (O(n)) plus the generated string (O(L)).
This approach expresses the mathematical definition of the sequence directly. It pairs naturally with problems that involve repeated transformations of previous results, a pattern commonly discussed in recursion. The tradeoff is extra stack usage and slightly more overhead than the iterative loop.
Recommended for interviews: The iterative approach is typically expected. It shows you can simulate the process efficiently using a single pass through the string. Mentioning the recursive formulation demonstrates understanding of the sequence definition, but the iterative solution signals stronger control over space usage and implementation details.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative String Simulation | O(n * L) | O(L) | Best general solution. Simple loop that builds each term sequentially with minimal overhead. |
| Recursive Construction | O(n * L) | O(n + L) | Useful when modeling the sequence definition directly or practicing recursion patterns. |