What is the time complexity of Special Binary String?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Special Binary String?

The space complexity is O(n^2).

What data structures are used to solve Special Binary String?

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

What are the key insights for solving Special Binary String?

Insight 1: A special binary string can be decomposed into smaller, nested special binary strings. The structure is inherently recursive. Insight 2: A special binary string always starts with '1' and ends with '0'. This characteristic allows us to identify candidate substrings.

Which companies ask LeetCode 761. Special Binary String in interviews?

This problem is frequently asked at Coursera, Grammarly, Nvidia, UKG.

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Special Binary String - LeetCode 761 Solution

Special Binary String - Complete Solution Guide

Special Binary String is LeetCode problem 761, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Special binary strings are binary strings with the following two properties: The number of 0 's is equal to the number of 1 's. Every prefix of the binary string has at least as many 1 's as 0 's. You are given a special binary string s . A move consists of choosing two consecutive, non-empty, special substrings of s , and swapping them. Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string. Return the lexicogra

Detailed Explanation

The problem defines a 'special binary string' as a string containing an equal number of '0's and '1's, where any prefix has at least as many '1's as '0's. The goal is to take a given special binary string `s` and rearrange its 'consecutive, non-empty, special substrings' (swapping them) to produce the lexicographically largest possible special binary string.

Solution Approach

The provided solution uses a recursive approach with memoization. It identifies special substrings within the input string, recursively sorts those substrings, and then concatenates them in descending order to achieve the lexicographically largest result. The memoization prevents redundant calculations for overlapping subproblems, improving efficiency.

Step-by-Step Algorithm

Step 1: Initialize memoization to store results of subproblems.
Step 2: Define a recursive function `solve(sub_s)` that takes a special binary string `sub_s` as input.
Step 3: In `solve(sub_s)`, check if the result for `sub_s` is already memoized. If so, return it directly.
Step 4: Iterate through the input string `sub_s`, maintaining a count of '1's and '0's. When the count returns to 0, it indicates the end of a special substring.
Step 5: Extract the middle part of the special substring (excluding the initial '1' and final '0'), and recursively call `solve` to find the largest special string for that middle part.
Step 6: Construct a new special substring by prepending '1' and appending '0' to the largest middle string.
Step 7: Collect all these reconstructed special substrings into a list of components.
Step 8: Sort the components list in descending lexicographical order.
Step 9: Concatenate the sorted components to form the largest special string for `sub_s`.
Step 10: Memoize the result for `sub_s` and return it.
Step 11: The initial call to `solve(s)` returns the final result.

Key Insights

Insight 1: A special binary string can be decomposed into smaller, nested special binary strings. The structure is inherently recursive.
Insight 2: A special binary string always starts with '1' and ends with '0'. This characteristic allows us to identify candidate substrings.
Insight 3: To maximize the lexicographical order, we need to sort the special substrings in descending order.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Recursion.

Companies

Asked at: Coursera, Grammarly, Nvidia, UKG.

Special Binary String | LeetCode 761

Hard

String Recursion

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

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

class Solution {
    private Map<String, String> memo = new HashMap<>();

    public String makeLargestSpecial(String s) {
        return solve(s);
    }

    private String solve(String sub_s) {
        if (memo.containsKey(sub_s)) {
            return memo.get(sub_s);
        }

        List<String> components = new ArrayList<>();
        int count = 0;
        int start = 0;
        for (int i = 0; i < sub_s.length(); i++) {
            char char_ = sub_s.charAt(i);
            if (char_ == '1') {
                count += 1;
            } else {
                count -= 1;
            }

            if (count == 0) {
                String middle = sub_s.substring(start + 1, i);
                String largest_middle = solve(middle);
                components.add("1" + largest_middle + "0");
                start = i + 1;
            }
        }

        Collections.sort(components, Collections.reverseOrder());

        String result = String.join("", components);
        memo.put(sub_s, result);
        return result;
    }
}

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Problem Info

Description

Problem 761: Special Binary String - Detailed explanation and solution approaches.

Topics

String Recursion

Companies

CourseraGrammarlyNvidiaUKG

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:761

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

11 steps

1
Step 1: Initialize memoization to store results of subproblems.
2
Step 2: Define a recursive function `solve(sub_s)` that takes a special binary string `sub_s` as input.
3
Step 3: In `solve(sub_s)`, check if the result for `sub_s` is already memoized. If so, return it directly.
4
Step 4: Iterate through the input string `sub_s`, maintaining a count of '1's and '0's. When the count returns to 0, it indicates the end of a special substring.
5
Step 5: Extract the middle part of the special substring (excluding the initial '1' and final '0'), and recursively call `solve` to find the largest special string for that middle part.
6
Step 6: Construct a new special substring by prepending '1' and appending '0' to the largest middle string.
7
Step 7: Collect all these reconstructed special substrings into a list of components.
8
Step 8: Sort the components list in descending lexicographical order.
9
Step 9: Concatenate the sorted components to form the largest special string for `sub_s`.
10
Step 10: Memoize the result for `sub_s` and return it.
11
Step 11: The initial call to `solve(s)` returns the final result.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity stems from the recursive nature of the algorithm and the sorting step. Each call to solve() iterates through the substring. The sorting is done on substrings, contributing an O(n log n) part, combined with the O(n) loop makes it O(n^2). The space complexity arises from storing the memoized results and storing list of strings, each up to length n.

📖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

Iterative Approach using Stack

An iterative approach could use a stack to keep track of the '1's and '0's balance. When the balance reaches 0, a special substring is identified. However, the lexicographical sorting aspect will still require a sorting step after extracting all substrings.

Time: O(n log n) (dominated by sorting)Space: O(n)

Trade-offs: While potentially slightly more efficient than the recursive memoized solution in terms of constant factors, the overall time complexity remains the same. The stack-based approach might be harder to implement and understand compared to the recursive solution.

✨

Key Insights

1Insight 1: A special binary string can be decomposed into smaller, nested special binary strings. The structure is inherently recursive.
2Insight 2: A special binary string always starts with '1' and ends with '0'. This characteristic allows us to identify candidate substrings.
3Insight 3: To maximize the lexicographical order, we need to sort the special substrings in descending order.

📚

Key Terms Explained

For Students

Recursion

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

String

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

Edge Cases

Edge case 1: Empty string. While the problem statement specifies that the length of the string will be greater than or equal to 1, it's a good practice to consider the empty string case. The recursion base case implicitly handles this by returning an empty string, avoiding any errors.
Edge case 2: Simple special strings like "10". The algorithm correctly identifies the string itself as a special string and returns it directly.
Edge case 3: Highly nested special strings like "111000". The recursive nature of the algorithm handles such cases by recursively decomposing the string into its special substrings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly identifying special substrings. Ensuring that every prefix has at least as many 1's as 0's is crucial.

✗Common mistake 2: Not sorting the special substrings in descending lexicographical order, which leads to a non-optimal result.

✗Common mistake 3: Forgetting to memoize, which can lead to exponential time complexity for certain inputs.

✗Common mistake 4: Not handling the base cases correctly can cause stack overflow errors or wrong results.

✅

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 algorithms. Special binary strings, or variations of them, might be used in data compression techniques to represent structured data, and rearranging these strings could lead to better compression ratios.

🏦Real-world scenario 2: Network packet optimization. In networking, packets might be represented as binary strings. Optimizing the order of certain sub-sequences (special substrings) within a packet could potentially lead to improved network performance.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Special Binary String"?

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

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