What is the time complexity of Count and Say?

The optimal solution has a time complexity of O(n*k).

What data structures are used to solve Count and Say?

This problem uses String. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count and Say?

Insight 1: Understanding the recursive definition is crucial. Each term in the sequence is generated from the previous term using run-length encoding. Insight 2: The problem can be solved iteratively by building the sequence from the first term up to the nth term.

Which companies ask LeetCode 38. Count and Say in interviews?

This problem is frequently asked at Adobe, Akuna Capital, Amazon, Apple, Bloomberg and 9 more companies.

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Count and Say - LeetCode 38 Solution

Count and Say - Complete Solution Guide

Count and Say is LeetCode problem 38, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 of countAndSay(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 "

Detailed Explanation

The "Count and Say" problem involves generating a sequence of strings based on a recursive rule. The first string in the sequence is "1". To generate the subsequent string, you describe the previous string by counting consecutive identical digits. For instance, if the previous string is "11", you say "two 1s", which becomes "21". If the previous string is "21", you say "one 2, one 1", which becomes "1211". The problem asks you to write a function that takes an integer `n` as input and returns the `n`-th string in this sequence. The constraints are that `n` will be between 1 and 30, inclusive.

Solution Approach

The provided solutions use an iterative approach to generate the count-and-say sequence. They start with the base case "1" and iteratively build the next string in the sequence `n-1` times. In each iteration, the code traverses the previous string, counts consecutive occurrences of the same character, and appends the count and the character to the next string being built. A StringBuilder (Java), vector of strings (C++), or list (Python) is used to efficiently construct the next string.

Step-by-Step Algorithm

Step 1: Initialize the first string as "1".
Step 2: Iterate from 2 to n. In each iteration, generate the next string based on the previous string.
Step 3: Inside the inner loop, iterate through the characters of the previous string.
Step 4: Count the consecutive occurrences of the same character.
Step 5: Append the count and the character to the new string.
Step 6: After iterating through the entire previous string, update the previous string with the newly generated string.
Step 7: After the outer loop completes, return the final generated string.

Key Insights

Insight 1: Understanding the recursive definition is crucial. Each term in the sequence is generated from the previous term using run-length encoding.
Insight 2: The problem can be solved iteratively by building the sequence from the first term up to the nth term.
Insight 3: String manipulation is key to building each new term in the sequence. We iterate through the previous string, count consecutive characters, and build the new string.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(k)

Topics

This problem involves: String.

Companies

Asked at: Adobe, Akuna Capital, Amazon, Apple, Bloomberg, ConsultAdd, Meta, Microsoft, PayPal, Pinterest, Wells Fargo, Wix, Yahoo, Zoho.

Count and Say | LeetCode 38

Medium

String

Time: O(n*k)Space: O(k)

Solution Code

Available in:

class Solution {
    public String countAndSay(int n) {
        if (n == 1) {
            return "1";
        }

        String current_str = "1";
        for (int _ = 0; _ < n - 1; _++) {
            StringBuilder next_str_parts = new StringBuilder();
            int i = 0;
            while (i < current_str.length()) {
                int count = 1;
                while (i + 1 < current_str.length() && current_str.charAt(i) == current_str.charAt(i + 1)) {
                    count += 1;
                    i += 1;
                }

                next_str_parts.append(String.valueOf(count));
                next_str_parts.append(current_str.charAt(i));
                i += 1;
            }

            current_str = next_str_parts.toString();
        }

        return current_str;
    }
}

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Problem Info

Description

Problem 38: Count and Say - Detailed explanation and solution approaches.

Topics

String

Companies

AdobeAkuna CapitalAmazonAppleBloombergConsultAddMetaMicrosoft

Stats

Difficulty:Medium

Languages:4

Companies:14

Problem #:38

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize the first string as "1".
2
Step 2: Iterate from 2 to n. In each iteration, generate the next string based on the previous string.
3
Step 3: Inside the inner loop, iterate through the characters of the previous string.
4
Step 4: Count the consecutive occurrences of the same character.
5
Step 5: Append the count and the character to the new string.
6
Step 6: After iterating through the entire previous string, update the previous string with the newly generated string.
7
Step 7: After the outer loop completes, return the final generated string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*k)

💾

Space Complexity

O(k)

📝

Why?

The outer loop runs `n-1` times. The inner loop iterates through the characters of the string generated in the previous iteration. The length of the string grows with each iteration, and k represents the average string length. The space complexity is determined by the space needed to store the longest string in the sequence, which is also proportional to k.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

The problem can also be solved recursively. The base case is n=1, which returns "1". For n>1, recursively call the function with n-1 and then generate the next string based on the returned string.

Time: O(n*k)Space: O(n*k) due to call stack and storing intermediate strings.

Trade-offs: The recursive approach is more intuitive but can be less efficient due to the overhead of function calls and potentially higher space complexity. The iterative approach is generally preferred for this problem due to better space usage.

✨

Key Insights

1Insight 1: Understanding the recursive definition is crucial. Each term in the sequence is generated from the previous term using run-length encoding.
2Insight 2: The problem can be solved iteratively by building the sequence from the first term up to the nth term.
3Insight 3: String manipulation is key to building each new term in the sequence. We iterate through the previous string, count consecutive characters, and build the new string.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: n = 1. The solution correctly handles this case by returning "1" directly.
Edge case 2: Larger values of n will result in longer string lengths. The solutions are designed to handle strings of varying length, limited by the problem constraints (n <= 30).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the base case (n=1) correctly.

✗Common mistake 2: Incorrectly counting consecutive characters, leading to wrong strings in the sequence.

✗Common mistake 3: Not handling the string building properly. Especially for C/C++, dynamic memory allocation for the intermediate strings can be tricky. It's essential to allocate enough memory and deallocate it properly to avoid memory leaks.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. Run-length encoding, the core concept behind this problem, is a basic form of lossless data compression, used in image and video compression.

🏦Real-world scenario 2: Pattern recognition. The generated sequences can be used as test cases or benchmarks for pattern recognition algorithms.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count and Say"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?