What is the time complexity of Decode String?

The optimal solution has a time complexity of O(N*M).

What is the space complexity of Decode String?

The space complexity is O(N*M).

What data structures are used to solve Decode String?

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

What are the key insights for solving Decode String?

Insight 1: The nested nature of the encoded string suggests the use of a stack to keep track of the current string and repetition count. Insight 2: When encountering a ']', we need to retrieve the previous string and repetition count from the stack to decode the current segment.

Which companies ask LeetCode 394. Decode String in interviews?

This problem is frequently asked at Activision, Agoda, ByteDance, Cisco, Compass and 24 more companies.

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Decode String - LeetCode 394 Solution

Decode String - Complete Solution Guide

Decode String is LeetCode problem 394, 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 an encoded string, return its decoded string. The encoding rule is: k[encoded_string] , where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k . For exam

Detailed Explanation

The problem asks us to decode a string that follows a specific encoding rule: `k[encoded_string]`, where `encoded_string` is repeated `k` times. The input string will consist of lowercase English letters, digits, and square brackets. The integer `k` represents the number of times the `encoded_string` inside the brackets should be repeated. The goal is to return the fully decoded string.

Solution Approach

The provided solutions use a stack-based approach to handle the nested structure of the encoded string. The algorithm iterates through the string character by character. When a digit is encountered, it accumulates it to form the repetition count. When an opening bracket '[' is encountered, the current string and repetition count are pushed onto the stack. When a closing bracket ']' is encountered, the algorithm pops the previous string and repetition count from the stack, repeats the current string by the specified number, and appends it to the previous string. Finally, when a lowercase letter is encountered, it's appended to the current string.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store pairs of (string, count), an integer to store the current number, and a string to store the current decoded string segment.
Step 2: Iterate through the input string character by character.
Step 3: If the current character is a digit, update the current number by multiplying the existing number by 10 and adding the digit.
Step 4: If the current character is an opening bracket '[', push the current string and number onto the stack, reset the current string to an empty string, and reset the current number to 0.
Step 5: If the current character is a closing bracket ']', pop the last string and number from the stack. Repeat the current string 'number' times and concatenate it with the popped string. Update the current string with this concatenated string.
Step 6: If the current character is a letter, append it to the current string.
Step 7: After iterating through the entire string, the final decoded string will be stored in the current string. Return the current string.

Key Insights

Insight 1: The nested nature of the encoded string suggests the use of a stack to keep track of the current string and repetition count.
Insight 2: When encountering a ']', we need to retrieve the previous string and repetition count from the stack to decode the current segment.
Insight 3: Numbers can be multi-digit, so accumulate digits until a non-digit character is encountered.

Complexity Analysis

Time Complexity: O(N*M)

Space Complexity: O(N*M)

Topics

This problem involves: String, Stack, Recursion.

Companies

Asked at: Activision, Agoda, ByteDance, Cisco, Compass, Coupang, Flexport, Geico, Goldman Sachs, Hertz, Huawei, Intuit, Mountblue, Nutanix, Ola Cabs, Ozon, Roku, Salesforce, Splunk, Sumo Logic, Tencent, Tesco, Tinkoff, Walmart Labs, Wix, Yelp, Zoho, eBay, razorpay.

Decode String | LeetCode 394

Medium

String Stack Recursion

Time: O(N*M)Space: O(N*M)

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public String decodeString(String s) {
        Stack<Object[]> stack = new Stack<>();
        int currentNum = 0;
        String currentString = "";

        for (char charAt : s.toCharArray()) {
            if (Character.isDigit(charAt)) {
                currentNum = currentNum * 10 + (charAt - '0');
            } else if (charAt == '[') {
                stack.push(new Object[]{currentString, currentNum});
                currentString = "";
                currentNum = 0;
            } else if (charAt == ']') {
                Object[] prev = stack.pop();
                String prevString = (String) prev[0];
                int num = (int) prev[1];
                StringBuilder repeatedString = new StringBuilder();
                for (int i = 0; i < num; i++) {
                    repeatedString.append(currentString);
                }
                currentString = prevString + repeatedString.toString();
            } else {
                currentString += charAt;
            }
        }

        return currentString;
    }
}

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Problem Info

Description

Problem 394: Decode String - Detailed explanation and solution approaches.

Topics

String Stack Recursion

Companies

ActivisionAgodaByteDanceCiscoCompassCoupangFlexportGeico

Stats

Difficulty:Medium

Languages:4

Companies:29

Problem #:394

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 an empty stack to store pairs of (string, count), an integer to store the current number, and a string to store the current decoded string segment.
2
Step 2: Iterate through the input string character by character.
3
Step 3: If the current character is a digit, update the current number by multiplying the existing number by 10 and adding the digit.
4
Step 4: If the current character is an opening bracket '[', push the current string and number onto the stack, reset the current string to an empty string, and reset the current number to 0.
5
Step 5: If the current character is a closing bracket ']', pop the last string and number from the stack. Repeat the current string 'number' times and concatenate it with the popped string. Update the current string with this concatenated string.
6
Step 6: If the current character is a letter, append it to the current string.
7
Step 7: After iterating through the entire string, the final decoded string will be stored in the current string. Return the current 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*M)

💾

Space Complexity

O(N*M)

📝

Why?

The time complexity arises from the iterations and string concatenations. In a scenario like '100[a]', the string 'a' will be repeated 100 times. Similarly, the space complexity comes from storing intermediate strings on the stack and creating new strings during the concatenation process. Deep nesting will result in many entries on 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 can also be used to solve this problem. It involves finding the innermost encoded string, decoding it recursively, and then proceeding outwards. This involves scanning the string for matching '[' and ']' brackets and recursing on the substring between them.

Time: O(N*M), where N is the length of the string, and M is the repetition factor.Space: O(N), where N is the maximum depth of the recursion, which is proportional to the level of nesting.

Trade-offs: The recursive approach might be more intuitive for some, but it can be less efficient due to function call overhead and the potential for stack overflow in very deeply nested cases. The iterative stack-based approach is generally more efficient.

✨

Key Insights

1Insight 1: The nested nature of the encoded string suggests the use of a stack to keep track of the current string and repetition count.
2Insight 2: When encountering a ']', we need to retrieve the previous string and repetition count from the stack to decode the current segment.
3Insight 3: Numbers can be multi-digit, so accumulate digits until a non-digit character is encountered.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

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 input string. The algorithm should return an empty string, which is handled correctly as the loop won't execute.
Edge case 2: Input string with no brackets or digits (e.g., "abc"). The algorithm will simply return the input string itself.
Edge case 3: Deeply nested brackets (e.g., "2[a3[b4[c]]]"). The stack will handle the nesting correctly, ensuring correct decoding.
Edge case 4: Large repetition counts (e.g., "300[abc]"). The algorithm will handle it, though it might be slow for extremely large numbers due to the string concatenation.
Edge case 5: Consecutive digits (e.g., "12[a]"). The algorithm correctly parses this by accumulating digits using multiplication by 10.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to reset the `current_num` after pushing onto the stack or after processing a closing bracket. This can lead to incorrect repetition counts.

✗Common mistake 2: Incorrectly handling multi-digit numbers. It's crucial to accumulate digits until a non-digit character is encountered.

✗Common mistake 3: Not handling the case where the input string has characters outside the encoded brackets. Ensure to add the characters to current string outside of bracket operations

✅

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. Although this specific encoding is simple, the general concept of repeating patterns is used in more sophisticated compression algorithms.

🏦Real-world scenario 2: Text generation. This algorithm can be seen as a simplified form of generative models where a short code is expanded into a longer text based on predefined rules.

💡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: July 5, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

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 problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

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 "Decode String"?

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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