What is the time complexity of Find the Original Typed String I?

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

What data structures are used to solve Find the Original Typed String I?

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

What are the key insights for solving Find the Original Typed String I?

Insight 1: The problem involves identifying consecutive repeating characters in the string. The number of possible original strings is directly related to the length of these repeating sequences. Insight 2: A simple iterative approach is sufficient to solve the problem. We can traverse the string once, counting consecutive repetitions.

Which companies ask LeetCode 3330. Find the Original Typed String I in interviews?

This problem is frequently asked at Lowe's.

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Find the Original Typed String I - LeetCode 3330 Solution

Find the Original Typed String I - Complete Solution Guide

Find the Original Typed String I is LeetCode problem 3330, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice is attempting to type a specific string on her computer. However, she tends to be clumsy and may press a key for too long, resulting in a character being typed multiple times. Although Alice tried to focus on her typing, she is aware that she may still have done this at most once . You are given a string word , which represents the final output displayed on Alice's screen. Return the total number of possible original strings that Alice might have intended to type. Example 1: Input: word =

Detailed Explanation

The problem asks to find the number of possible original strings that Alice could have typed, given a final output string where she might have pressed a key multiple times consecutively. The input is a string `word` representing the final output. The output is an integer representing the count of possible original strings. Alice could have pressed a key repeatedly at most once in the entire string. The constraint is that the input string's length is between 1 and 100 characters, inclusive, and contains only lowercase English letters.

Solution Approach

The solution uses a single pass through the input string. It iterates character by character, identifying consecutive sequences of identical characters. For each such sequence, if its length is greater than 1, it means Alice pressed the key multiple times. The number of possible original strings for that sequence is equal to the length of the sequence minus 1 (original + one less repetition). These counts for all such sequences are summed to get the final answer.

Step-by-Step Algorithm

Step 1: Initialize a `count` variable to 1 (representing the original string itself).
Step 2: Iterate through the string. For each character, check if it is the same as the next character.
Step 3: If it is the same, find the length of the consecutive sequence of identical characters.
Step 4: If the length of the sequence is greater than 1, add `length - 1` to the `count`.
Step 5: Continue iterating until the end of the string.
Step 6: Return the final `count`.

Key Insights

Insight 1: The problem involves identifying consecutive repeating characters in the string. The number of possible original strings is directly related to the length of these repeating sequences.
Insight 2: A simple iterative approach is sufficient to solve the problem. We can traverse the string once, counting consecutive repetitions.
Insight 3: Handling the edge case of a string with no consecutive repeating characters (the count remains 1) is crucial.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Lowe's.

Find the Original Typed String I | LeetCode 3330

Easy

String

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int possibleStringCount(String word) {
        int count = 1;
        int n = word.length();
        for (int i = 0; i < n - 1; i++) {
            if (word.charAt(i) == word.charAt(i + 1)) {
                int j = i + 1;
                while (j < n && word.charAt(i) == word.charAt(j)) {
                    j++;
                }
                count += (j - i - 1);
                i = j - 1;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 3330: Find the Original Typed String I - Detailed explanation and solution approaches.

Topics

String

Companies

Lowe's

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3330

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

6 steps

1
Step 1: Initialize a `count` variable to 1 (representing the original string itself).
2
Step 2: Iterate through the string. For each character, check if it is the same as the next character.
3
Step 3: If it is the same, find the length of the consecutive sequence of identical characters.
4
Step 4: If the length of the sequence is greater than 1, add `length - 1` to the `count`.
5
Step 5: Continue iterating until the end of the string.
6
Step 6: Return the final `count`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the string only once. No auxiliary data structures that scale with the input size are used. Therefore, both time and space complexity are linear with respect to the input size, but the space complexity is constant because the extra space used is independent of the input size.

📖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

A recursive solution could be implemented to process the string character by character. It would recursively explore the possibilities of removing one or more consecutive duplicate characters.

Time: O(n*m), where n is the length of the input and m is the maximum number of consecutive identical characters (worst case).Space: O(n) due to recursive call stack.

Trade-offs: The recursive approach is less efficient than the iterative approach due to the overhead of recursive calls and potentially larger space complexity. The iterative approach is preferred for its simplicity and efficiency.

✨

Key Insights

1Insight 1: The problem involves identifying consecutive repeating characters in the string. The number of possible original strings is directly related to the length of these repeating sequences.
2Insight 2: A simple iterative approach is sufficient to solve the problem. We can traverse the string once, counting consecutive repetitions.
3Insight 3: Handling the edge case of a string with no consecutive repeating characters (the count remains 1) is crucial.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: The input string contains no consecutive repeating characters. The solution correctly handles this by returning 1.
Edge case 2: The input string contains only one type of character, e.g., "aaaaa". The solution correctly calculates the number of possible strings as the length of the string minus 1 (4 in this case).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of possible original strings for a sequence of repeating characters (e.g., off by one error).

✗Common mistake 2: Not handling the edge case of a string with no consecutive repeating characters correctly.

✅

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 cleaning and normalization. This algorithm could be used to identify and correct potential typing errors in data entry systems.

🏦Real-world scenario 2: Text compression. Identifying repeated sequences of characters is a fundamental step in many text compression algorithms. While this specific problem doesn't directly solve compression, it shows the underlying pattern recognition relevant to compression 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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning String. It's a great foundation for more complex problems.

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: August 30, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Find the Original Typed String I"?

For this easy-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.

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