What is the time complexity of Minimum Number of Swaps to Make the String Balanced?

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

What data structures are used to solve Minimum Number of Swaps to Make the String Balanced?

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

What are the key insights for solving Minimum Number of Swaps to Make the String Balanced?

Insight 1: The number of opening and closing brackets are equal, meaning a balanced state exists. Insight 2: The problem can be solved by tracking the imbalance between opening and closing brackets as we iterate through the string.

Which companies ask LeetCode 1963. Minimum Number of Swaps to Make the String Balanced in interviews?

This problem is frequently asked at Expedia, Microstrategy, Nutanix, PayPal, Twilio.

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Minimum Number of Swaps to Make the String Balanced - LeetCode 1963 Solution

Minimum Number of Swaps to Make the String Balanced - Complete Solution Guide

Minimum Number of Swaps to Make the String Balanced is LeetCode problem 1963, 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 0-indexed string s of even length n . The string consists of exactly n / 2 opening brackets '[' and n / 2 closing brackets ']' . A string is called balanced if and only if: It is the empty string, or It can be written as AB , where both A and B are balanced strings, or It can be written as [C] , where C is a balanced string. You may swap the brackets at any two indices any number of times. Return the minimum number of swaps to make s balanced . Example 1: Input: s = "][][" Output

Detailed Explanation

The problem requires finding the minimum number of swaps needed to balance a string `s` containing an equal number of opening brackets '[' and closing brackets ']'. A balanced string can be empty, formed by concatenating two balanced strings, or enclosed within square brackets where the enclosed string is balanced. The string initially has mismatched brackets and needs to be balanced using swaps between any two indices.

Solution Approach

The solution iterates through the string, maintaining a `balance` counter. This counter is decremented when an opening bracket '[' is encountered and incremented when a closing bracket ']' is encountered. `max_balance` keeps track of the maximum value the `balance` counter reaches during the iteration. The final answer is derived from `max_balance`. The intuition is that `max_balance` indicates the maximum number of extra closing brackets encountered before their corresponding opening brackets. We need to swap some closing brackets with opening brackets that appear later in the string to correct the imbalance. The formula (max_balance + 1) // 2 calculates the minimum number of swaps required to achieve balance.

Step-by-Step Algorithm

Step 1: Initialize `balance` and `max_balance` to 0.
Step 2: Iterate through the string `s` from left to right.
Step 3: If the current character is '[', decrement the `balance` counter.
Step 4: If the current character is ']', increment the `balance` counter.
Step 5: Update `max_balance` to be the maximum of its current value and the current `balance`.
Step 6: After the loop finishes, return `(max_balance + 1) // 2`.
Step 7: Integer division is used to determine the minimum number of swaps needed.

Key Insights

Insight 1: The number of opening and closing brackets are equal, meaning a balanced state exists.
Insight 2: The problem can be solved by tracking the imbalance between opening and closing brackets as we iterate through the string.
Insight 3: The maximum imbalance encountered during the iteration indicates the minimum number of swaps needed.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Two Pointers, String, Stack, Greedy.

Companies

Asked at: Expedia, Microstrategy, Nutanix, PayPal, Twilio.

Minimum Number of Swaps to Make the String Balanced | LeetCode 1963

Medium

Two Pointers String Stack Greedy

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

Solution Code

Available in:

class Solution {
    public int minSwaps(String s) {
        int balance = 0;
        int max_balance = 0;
        for (char char_s : s.toCharArray()) {
            if (char_s == '[') {
                balance -= 1;
            } else {
                balance += 1;
            }
            max_balance = Math.max(max_balance, balance);
        }
        
        return (max_balance + 1) / 2;
    }
}

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Problem Info

Description

Problem 1963: Minimum Number of Swaps to Make the String Balanced - Detailed explanation and solution approaches.

Topics

Two Pointers String Stack Greedy

Companies

ExpediaMicrostrategyNutanixPayPalTwilio

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1963

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 `balance` and `max_balance` to 0.
2
Step 2: Iterate through the string `s` from left to right.
3
Step 3: If the current character is '[', decrement the `balance` counter.
4
Step 4: If the current character is ']', increment the `balance` counter.
5
Step 5: Update `max_balance` to be the maximum of its current value and the current `balance`.
6
Step 6: After the loop finishes, return `(max_balance + 1) // 2`.
7
Step 7: Integer division is used to determine the minimum number of swaps needed.

💡

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 time complexity is determined by the single loop iterating through the string, making it linear with respect to the string's length. The space complexity is constant because only a fixed number of integer variables are used, irrespective of the string's 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

Stack-based approach

Use a stack to keep track of unmatched opening brackets. When a closing bracket is encountered, check if the stack is empty. If not, pop an opening bracket. If it is empty, it means we need to swap this closing bracket with a later opening bracket. This approach also requires keeping track of the swap count.

Time: O(n)Space: O(n) in the worst case (e.g., '[[[[[[').

Trade-offs: The stack-based approach has higher space complexity compared to the O(1) space complexity of the optimal solution. The provided solution is better because it's more space-efficient.

✨

Key Insights

1Insight 1: The number of opening and closing brackets are equal, meaning a balanced state exists.
2Insight 2: The problem can be solved by tracking the imbalance between opening and closing brackets as we iterate through the string.
3Insight 3: The maximum imbalance encountered during the iteration indicates the minimum number of swaps needed.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Stack

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty string: The algorithm will correctly return 0 because `balance` and `max_balance` will remain 0.
Edge case 2: Already balanced string (e.g., '[]'): The algorithm will correctly return 0 because `max_balance` will be 0.
Edge case 3: String with severely unbalanced closing brackets at the beginning (e.g., ']]]]'): The algorithm will accurately compute the number of swaps needed based on `max_balance`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the significance of `max_balance`. It represents the maximum excess of closing brackets encountered before their corresponding opening brackets.

✗Common mistake 2: Forgetting to add 1 before dividing by 2 in the final calculation. `(max_balance+1)//2` is crucial for correctly determining the minimum number of swaps.

✗Common mistake 3: Confusing the problem with finding the minimum number of operations to balance any arbitrary string with potentially unequal counts of '[' and ']'.

✅

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: Balancing delimiters in programming languages (parentheses, curly braces, etc.). While this specific problem has equal counts, the concept of tracking balance is crucial for validating code syntax.

🏦Real-world scenario 2: Optimizing resource allocation where resources need to be paired (e.g., allocating servers to requests). Minimizing the number of reallocations can be seen as analogous to minimizing swaps in this problem.

💡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

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

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

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

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: November 11, 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

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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.

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 "Minimum Number of Swaps to Make the String Balanced"?

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

🤖 Need help with this problem?