What is the time complexity of Check If Word Is Valid After Substitutions?

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

What data structures are used to solve Check If Word Is Valid After Substitutions?

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

What are the key insights for solving Check If Word Is Valid After Substitutions?

Insight 1: The problem can be efficiently solved using a stack. When encountering an 'a' or 'b', push it onto the stack. When encountering a 'c', check if the top two elements of the stack are 'b' and 'a' respectively. If so, pop them off the stack; otherwise, the string is invalid. Insight 2: The order of 'a', 'b', and 'c' matters. 'a' must come before 'b' which must come before 'c'. This suggests a last-in, first-out (LIFO) data structure like a stack is appropriate.

Which companies ask LeetCode 1003. Check If Word Is Valid After Substitutions in interviews?

This problem is frequently asked at Nutanix.

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Check If Word Is Valid After Substitutions - LeetCode 1003 Solution

Check If Word Is Valid After Substitutions - Complete Solution Guide

Check If Word Is Valid After Substitutions is LeetCode problem 1003, 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

Given a string s , determine if it is valid . A string s is valid if, starting with an empty string t = "" , you can transform t into s after performing the following operation any number of times : Insert string "abc" into any position in t . More formally, t becomes t left + "abc" + t right , where t == t left + t right . Note that t left and t right may be empty . Return true if s is a valid string, otherwise, return false . Example 1: Input: s = "aabcbc" Output: true Explanation: "" -> " abc

Detailed Explanation

The problem asks us to determine if a given string `s` is valid. A string is considered valid if it can be constructed by repeatedly inserting the string "abc" into any position of an initially empty string. Essentially, we need to check if we can reduce the string `s` back to an empty string by repeatedly removing "abc" from it.

Solution Approach

The solution uses a stack to simulate the process of removing "abc" from the string. The algorithm iterates through the input string character by character. If the current character is 'a' or 'b', it's pushed onto the stack. If the current character is 'c', the algorithm checks if the top two elements of the stack are 'b' and 'a' respectively. If they are, it pops both 'b' and 'a' from the stack, effectively removing an "abc" sequence. If the top elements are not 'b' and 'a', or if the stack doesn't contain enough elements, the string is invalid. After processing all characters, the algorithm checks if the stack is empty. An empty stack means the entire string has been reduced to an empty string, indicating the string is valid.

Step-by-Step Algorithm

Step 1: Initialize an empty stack (or array used as a stack).
Step 2: Iterate through the input string `s` character by character.
Step 3: If the current character is 'a' or 'b', push it onto the stack.
Step 4: If the current character is 'c':
Step 4.1: Check if the stack has at least two elements.
Step 4.2: If not, the string is invalid, return false.
Step 4.3: Pop the top element from the stack. If it's not 'b', the string is invalid, return false.
Step 4.4: Pop the next element from the stack. If it's not 'a', the string is invalid, return false.
Step 5: After processing all characters, check if the stack is empty.
Step 6: If the stack is empty, the string is valid, return true. Otherwise, the string is invalid, return false.

Key Insights

Insight 1: The problem can be efficiently solved using a stack. When encountering an 'a' or 'b', push it onto the stack. When encountering a 'c', check if the top two elements of the stack are 'b' and 'a' respectively. If so, pop them off the stack; otherwise, the string is invalid.
Insight 2: The order of 'a', 'b', and 'c' matters. 'a' must come before 'b' which must come before 'c'. This suggests a last-in, first-out (LIFO) data structure like a stack is appropriate.
Insight 3: If, at the end of processing the string, the stack is empty, it means the entire string was successfully reduced to an empty string via "abc" removals, making it a valid string. If the stack is not empty, it means some characters could not be matched, and the string is invalid.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack.

Companies

Asked at: Nutanix.

Check If Word Is Valid After Substitutions | LeetCode 1003

Medium

String Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<>();
        for (char char_ : s.toCharArray()) {
            if (char_ == 'c') {
                if (stack.isEmpty() || stack.pop() != 'b') {
                    return false;
                }
                if (stack.isEmpty() || stack.pop() != 'a') {
                    return false;
                }
            } else {
                stack.push(char_);
            }
        }

        return stack.isEmpty();
    }
}

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Problem Info

Description

Problem 1003: Check If Word Is Valid After Substitutions - Detailed explanation and solution approaches.

Topics

String Stack

Companies

Nutanix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1003

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

10 steps

1
Step 1: Initialize an empty stack (or array used as a stack).
2
Step 2: Iterate through the input string `s` character by character.
3
Step 3: If the current character is 'a' or 'b', push it onto the stack.
4
Step 4: If the current character is 'c':
5
Step 4.1: Check if the stack has at least two elements.
6
Step 4.2: If not, the string is invalid, return false.
7
Step 4.3: Pop the top element from the stack. If it's not 'b', the string is invalid, return false.
8
Step 4.4: Pop the next element from the stack. If it's not 'a', the string is invalid, return false.
9
Step 5: After processing all characters, check if the stack is empty.
10
Step 6: If the stack is empty, the string is valid, return true. Otherwise, the string is invalid, return false.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because each character in the string is processed only once. The space complexity is also linear because, in the worst case, we might need to store almost all characters in the stack.

📖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 approach could involve searching for "abc" within the string, removing it, and recursively calling the function on the remaining string. The base case would be an empty string (valid) or a string that doesn't contain "abc" anymore (invalid).

Time: O(n^2) in the worst case, where n is the length of the string.Space: O(n) due to the recursive call stack.

Trade-offs: The recursive approach is generally less efficient than the stack-based approach due to the overhead of function calls and potential for repeated calculations. It might be more readable for some, but the stack approach is generally favored for better performance.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using a stack. When encountering an 'a' or 'b', push it onto the stack. When encountering a 'c', check if the top two elements of the stack are 'b' and 'a' respectively. If so, pop them off the stack; otherwise, the string is invalid.
2Insight 2: The order of 'a', 'b', and 'c' matters. 'a' must come before 'b' which must come before 'c'. This suggests a last-in, first-out (LIFO) data structure like a stack is appropriate.
3Insight 3: If, at the end of processing the string, the stack is empty, it means the entire string was successfully reduced to an empty string via "abc" removals, making it a valid string. If the stack is not empty, it means some characters could not be matched, and the string is invalid.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

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

Edge Cases

Edge case 1: Empty string. The code implicitly handles this since the for loop won't execute and the stack will be empty, which will lead to the correct result.
Edge case 2: Strings that don't contain 'c'. These will result in characters being pushed onto the stack, and the stack will not be empty at the end, correctly identifying the string as invalid.
Edge case 3: Strings starting with 'b' or 'c'. These immediately trigger the 'stack is empty' condition within the 'c' handling part, resulting in the correct invalid outcome.
Edge case 4: Strings like 'ab' or 'ba'. Stack would not be empty at the end, so it will return false, which is the expected behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the order of 'a', 'b', and 'c' correctly. Failing to check that 'a' comes before 'b' before 'c' will lead to incorrect results.

✗Common mistake 2: Not checking if the stack is empty before popping elements. This will cause a runtime error in most languages and can lead to incorrect results.

✗Common mistake 3: Forgetting to check if the stack is empty at the end of the process. If the stack is not empty, the string is invalid, even if all the 'c' characters had corresponding 'a' and 'b' characters at some point.

✅

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: Syntax validation for simplified programming languages or data formats where specific sequences are required to be valid.

🏦Real-world scenario 2: Data compression or encoding algorithms where certain patterns need to be identified and removed to reduce data size (though "abc" is a trivial example, the principle applies to more complex patterns).

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: September 12, 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

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Check If Word Is Valid After Substitutions"?

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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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