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

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

What data structures are used to solve Minimum Deletions to Make String Balanced?

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

What are the key insights for solving Minimum Deletions to Make String Balanced?

Insight 1: The problem can be solved using dynamic programming or a greedy approach by iterating through the string once. Insight 2: For each position in the string, we can choose to either delete the current character, or keep it. Keeping 'a' means deleting all 'b's before it. Keeping 'b' means deleting all 'a's after it. The optimal solution will explore one of these options.

Which companies ask LeetCode 1653. Minimum Deletions to Make String Balanced in interviews?

This problem is frequently asked at redbus.

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Minimum Deletions to Make String Balanced - LeetCode 1653 Solution

Q: How to solve Minimum Deletions to Make String Balanced?

The solution uses a greedy approach to find the minimum deletions. It iterates through the string, keeping track of the number of 'b's encountered so far (`b_count`). When it encounters an 'a', it calculates the number of deletions needed in two scenarios: 1) delete the current 'a', which would increment `deletions` by 1, or 2) delete all 'b's encountered before this 'a', which is equal to `b_count`. The algorithm chooses the minimum of these two options and updates `deletions` accordingly. If a character is 'b', we simply increment `b_count`. This approach avoids storing the entire string or any large intermediate values, leading to an efficient solution.

Minimum Deletions to Make String Balanced - Complete Solution Guide

Minimum Deletions to Make String Balanced is LeetCode problem 1653, 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 consisting only of characters 'a' and 'b' . You can delete any number of characters in s to make s balanced . s is balanced if there is no pair of indices (i,j) such that i < j and s[i] = 'b' and s[j]= 'a' . Return the minimum number of deletions needed to make s balanced . Example 1: Input: s = "aababbab" Output: 2 Explanation: You can either: Delete the characters at 0-indexed positions 2 and 6 ("aa b abb a b" -> "aaabbb"), or Delete the characters at 0-indexed pos

Detailed Explanation

The problem requires finding the minimum number of character deletions in a string `s` (containing only 'a' and 'b' characters) to make it 'balanced'. A string is balanced if there are no indices `i` and `j` such that `i < j` and `s[i] = 'b'` and `s[j] = 'a'`. In simpler terms, all 'b's must appear before all 'a's, or the string must consist only of 'a's or only of 'b's.

Solution Approach

The solution uses a greedy approach to find the minimum deletions. It iterates through the string, keeping track of the number of 'b's encountered so far (`b_count`). When it encounters an 'a', it calculates the number of deletions needed in two scenarios: 1) delete the current 'a', which would increment `deletions` by 1, or 2) delete all 'b's encountered before this 'a', which is equal to `b_count`. The algorithm chooses the minimum of these two options and updates `deletions` accordingly. If a character is 'b', we simply increment `b_count`. This approach avoids storing the entire string or any large intermediate values, leading to an efficient solution.

Step-by-Step Algorithm

Step 1: Initialize `deletions` to 0 and `b_count` to 0. These variables will store the minimum number of deletions and the count of 'b's, respectively.
Step 2: Iterate through the input string `s` character by character.
Step 3: For each character, check if it is 'a'.
Step 4: If the character is 'a', update `deletions` to the minimum of `deletions + 1` (deleting the current 'a') and `b_count` (deleting all 'b's encountered before). This chooses the optimal deletion strategy at this point.
Step 5: If the character is 'b', increment `b_count` by 1.
Step 6: After iterating through the entire string, return the final value of `deletions`.

Key Insights

Insight 1: The problem can be solved using dynamic programming or a greedy approach by iterating through the string once.
Insight 2: For each position in the string, we can choose to either delete the current character, or keep it. Keeping 'a' means deleting all 'b's before it. Keeping 'b' means deleting all 'a's after it. The optimal solution will explore one of these options.
Insight 3: We can maintain a count of 'b's encountered so far. When we encounter an 'a', we have two choices: either delete the 'a' or delete all the 'b's seen before it. We choose the option that results in fewer deletions.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Dynamic Programming, Stack.

Companies

Asked at: redbus.

Minimum Deletions to Make String Balanced | LeetCode 1653

Medium

String Dynamic Programming Stack

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

Solution Code

Available in:

class Solution {
    public int minimumDeletions(String s) {
        int deletions = 0;
        int b_count = 0;
        
        for (int i = 0; i < s.length(); i++) {
            char char_val = s.charAt(i);
            if (char_val == 'a') {
                deletions = Math.min(deletions + 1, b_count);
            } else {
                b_count += 1;
            }
        }
        
        return deletions;
    }
}

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Problem Info

Description

Problem 1653: Minimum Deletions to Make String Balanced - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Stack

Companies

redbus

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1653

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

6 steps

1
Step 1: Initialize `deletions` to 0 and `b_count` to 0. These variables will store the minimum number of deletions and the count of 'b's, respectively.
2
Step 2: Iterate through the input string `s` character by character.
3
Step 3: For each character, check if it is 'a'.
4
Step 4: If the character is 'a', update `deletions` to the minimum of `deletions + 1` (deleting the current 'a') and `b_count` (deleting all 'b's encountered before). This chooses the optimal deletion strategy at this point.
5
Step 5: If the character is 'b', increment `b_count` by 1.
6
Step 6: After iterating through the entire string, return the final value of `deletions`.

💡

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 linear because the algorithm processes each character in the string once. The space complexity is constant because only a fixed number of variables are used irrespective of the input 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

Dynamic Programming

A dynamic programming approach could be used where `dp[i]` represents the minimum number of deletions needed to balance the string up to index `i`. The recurrence relation would consider the two cases: delete the character at index `i`, or keep it and ensure the string up to `i` is balanced. This approach is less efficient than the greedy approach.

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

Trade-offs: The DP approach has the same time complexity as the greedy approach but requires more space (O(n) vs O(1)). The greedy approach is generally preferred for its simplicity and space efficiency.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming or a greedy approach by iterating through the string once.
2Insight 2: For each position in the string, we can choose to either delete the current character, or keep it. Keeping 'a' means deleting all 'b's before it. Keeping 'b' means deleting all 'a's after it. The optimal solution will explore one of these options.
3Insight 3: We can maintain a count of 'b's encountered so far. When we encounter an 'a', we have two choices: either delete the 'a' or delete all the 'b's seen before it. We choose the option that results in fewer deletions.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

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 provided code correctly handles the empty string case because the loop won't execute, and the initial value of `deletions` (0) will be returned.
Edge case 2: String containing only 'a's. The algorithm will iterate through the string, and for each 'a', `deletions` will be updated to `min(deletions + 1, 0)`, which means `deletions` will increment by 1 for each 'a'. This means the result will be the number of a's which is the correct number of deletions to get an empty (balanced) string. Although this could be viewed as deleting everything, the problem description allows for any number of characters to be deleted, which would include nothing being deleted, in which case the string is already balanced.
Edge case 3: String containing only 'b's. The algorithm will iterate through the string, and only increment `b_count`. Since there are no 'a's, `deletions` will remain 0, which is the correct answer, as the string is already balanced.
Edge case 4: String like 'ba'. In this case, b_count will be 1, and deletions will be min(deletions + 1, b_count) = min(1,1) = 1. The optimal solution is to delete either 'b' or 'a'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the `min` function or using the wrong logic to determine whether to delete the current character or the preceding 'b's.

✗Common mistake 2: Failing to handle edge cases such as an empty string or a string containing only 'a's or 'b's.

✗Common mistake 3: Using a less efficient algorithm like brute force or recursion which would lead to Time Limit Exceeded errors for larger input strings.

✅

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 cleaning. When processing log files or other textual data, you might need to remove characters to enforce a specific format or order.

🏦Real-world scenario 2: String validation. In input validation, a similar approach could be used to minimize modifications to a string to conform to a particular pattern (e.g., removing invalid characters to make a string a valid identifier).

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 4, 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 14

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

@TechLeadJul 7

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

@DevExpert92Jun 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 Deletions to Make String Balanced"?

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

🤖 Need help with this problem?