What is the time complexity of Evaluate the Bracket Pairs of a String?

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

What is the space complexity of Evaluate the Bracket Pairs of a String?

The space complexity is O(k + n).

What data structures are used to solve Evaluate the Bracket Pairs of a String?

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

What are the key insights for solving Evaluate the Bracket Pairs of a String?

Insight 1: Using a hash table (dictionary) to store the knowledge allows for efficient lookups of key-value pairs. Insight 2: Splitting the input string 's' by '(' simplifies the processing of each bracketed section.

Which companies ask LeetCode 1807. Evaluate the Bracket Pairs of a String in interviews?

This problem is frequently asked at Remitly.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Evaluate the Bracket Pairs of a String - LeetCode 1807 Solution

Evaluate the Bracket Pairs of a String - Complete Solution Guide

Evaluate the Bracket Pairs of a String is LeetCode problem 1807, 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

You are given a string s that contains some bracket pairs, with each pair containing a non-empty key. For example, in the string "(name)is(age)yearsold" , there are two bracket pairs that contain the keys "name" and "age" . You know the values of a wide range of keys. This is represented by a 2D string array knowledge where each knowledge[i] = [key i , value i ] indicates that key key i has a value of value i . You are tasked to evaluate all of the bracket pairs. When you evaluate a bracket pair

Detailed Explanation

The problem requires evaluating bracket pairs in a string 's' based on a 'knowledge' base. Each bracket pair contains a key. The task is to replace each bracket pair and its key with the corresponding value from the 'knowledge' base. If a key is not found in 'knowledge', it should be replaced with '?'. The 'knowledge' is a 2D array where each row contains a key-value pair. The string 's' consists of lowercase letters and round brackets. Keys are non-empty and unique within the knowledge base, and there are no nested brackets.

Solution Approach

The provided solution uses a hash table (dictionary) to store the 'knowledge' for quick lookups. It splits the input string 's' by '(' characters. It then iterates through the resulting parts. For each part, it splits it further by ')' to extract the key and the remaining part of the string. The key is then looked up in the hash table. If found, the corresponding value is appended to the result. Otherwise, '?' is appended. Finally, the remaining part of the string is appended to the result.

Step-by-Step Algorithm

Step 1: Create a hash table (dictionary) from the 'knowledge' list to store key-value pairs for efficient lookups.
Step 2: Split the input string 's' by the '(' character into a list of parts.
Step 3: Initialize an empty result string with the first part (the portion before the first '(').
Step 4: Iterate through the remaining parts (starting from the second part) of the split string.
Step 5: For each part, split it by the ')' character to extract the 'key' and the 'rest' of the string.
Step 6: Look up the 'key' in the hash table. If found, append its corresponding 'value' to the result string. If not found, append '?' to the result string.
Step 7: Append the 'rest' of the string (the part after the ')') to the result string.
Step 8: After processing all parts, return the final result string.

Key Insights

Insight 1: Using a hash table (dictionary) to store the knowledge allows for efficient lookups of key-value pairs.
Insight 2: Splitting the input string 's' by '(' simplifies the processing of each bracketed section.
Insight 3: Handling the cases where the key is present or absent in the knowledge base is crucial for correctness.

Complexity Analysis

Time Complexity: O(n + k)

Space Complexity: O(k + n)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Remitly.

Evaluate the Bracket Pairs of a String | LeetCode 1807

Medium

Array Hash Table String

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public String evaluate(String s, List<List<String>> knowledge) {
        Map<String, String> knowledge_dict = new HashMap<>();
        for (List<String> pair : knowledge) {
            knowledge_dict.put(pair.get(0), pair.get(1));
        }

        String[] parts = s.split("\\(");
        StringBuilder res = new StringBuilder(parts[0]);

        for (int i = 1; i < parts.length; i++) {
            String part = parts[i];
            String[] keyAndRest = part.split("\\)", 2);
            String key = keyAndRest[0];
            String rest = keyAndRest[1];

            res.append(knowledge_dict.getOrDefault(key, "?"));
            res.append(rest);
        }

        return res.toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1807: Evaluate the Bracket Pairs of a String - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

Remitly

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1807

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

8 steps

1
Step 1: Create a hash table (dictionary) from the 'knowledge' list to store key-value pairs for efficient lookups.
2
Step 2: Split the input string 's' by the '(' character into a list of parts.
3
Step 3: Initialize an empty result string with the first part (the portion before the first '(').
4
Step 4: Iterate through the remaining parts (starting from the second part) of the split string.
5
Step 5: For each part, split it by the ')' character to extract the 'key' and the 'rest' of the string.
6
Step 6: Look up the 'key' in the hash table. If found, append its corresponding 'value' to the result string. If not found, append '?' to the result string.
7
Step 7: Append the 'rest' of the string (the part after the ')') to the result string.
8
Step 8: After processing all parts, return the final result 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 + n)

📝

Why?

The time complexity is determined by the string splitting and hash table lookups. The space complexity is determined by the size of the hash table and the final result string.

📖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

Using regular expressions

Regular expressions could be used to find and replace the bracket pairs in the string. This approach might be more concise but potentially less efficient than the splitting approach, especially for large strings.

Time: O(n*m) where n is the string length and m is the regex matching timeSpace: O(n)

Trade-offs: Regular expressions can be more readable for simple patterns, but can become harder to maintain. Splitting-based approach might offer better performance and easier debugging.

✨

Key Insights

1Insight 1: Using a hash table (dictionary) to store the knowledge allows for efficient lookups of key-value pairs.
2Insight 2: Splitting the input string 's' by '(' simplifies the processing of each bracketed section.
3Insight 3: Handling the cases where the key is present or absent in the knowledge base is crucial for correctness.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: The input string 's' contains no bracket pairs. In this case, the solution correctly returns the original string 's'.
Edge case 2: The 'knowledge' base is empty. In this case, all bracket pairs will be replaced with '?', which the solution handles correctly.
Edge case 3: A key exists multiple times in the input string. The solution correctly handles this by replacing each occurrence independently based on the knowledge base.
Edge case 4: An empty key `()` exists in the input string. The code handles splitting based on `(` and `)`. The logic will still work because an empty key will not be found in the knowledge and replaced by `?`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where a key is not found in the 'knowledge' base, resulting in an incorrect replacement or an error.

✗Common mistake 2: Incorrectly splitting the string, leading to incorrect extraction of keys and values.

✗Common mistake 3: Memory leaks (especially in C) if dynamically allocated memory is not freed after use.

✅

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: Configuration file parsing. Many configuration files contain bracketed placeholders that need to be replaced with values from a configuration store.

🏦Real-world scenario 2: Template engines. Template engines use similar techniques to replace placeholders with dynamic content. The provided algorithm forms the basis of a simple template engine.

🎮Real-world scenario 3: Data anonymization. Bracket pairs can be used to mark Personally Identifiable Information (PII) and replace them with anonymized values from a lookup table.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: June 8, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Evaluate the Bracket Pairs of a String"?

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

🤖 Need help with this problem?