What is the time complexity of Maximum Number of Balloons?

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

What data structures are used to solve Maximum Number of Balloons?

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

What are the key insights for solving Maximum Number of Balloons?

**Efficient Character Frequency Tracking**: The initial step of constructing a hash map (dictionary) for `text` characters is crucial. It provides an O(1) average time complexity lookup for character counts, allowing for rapid assessment of available resources without needing repeated string scans. **Handling Asymmetrical Letter Requirements**: The word 'balloon' does not require all its letters with the same frequency ('b', 'a', 'n' once; 'l', 'o' twice). The solution correctly accounts for this by dividing the available counts of 'l' and 'o' by two, effectively converting their raw counts into 'pairs' that satisfy the word's specific structure.

Which companies ask LeetCode 1189. Maximum Number of Balloons in interviews?

This problem is frequently asked at Tesla, Wayfair.

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Maximum Number of Balloons - LeetCode 1189 Solution

Q: How to solve Maximum Number of Balloons?

The provided solution elegantly solves this by first creating a comprehensive inventory of all characters available in the input `text`. It accomplishes this using a hash map (or dictionary in Python), where each character from `text` becomes a key, and its corresponding value is the count of how many times it appears. This gives us an immediate, efficient lookup for any character's availability. Once this character inventory is built, the solution focuses on the specific characters needed for "balloon": 'b', 'a', 'l', 'o', 'n'. It gathers the counts for these five letters from our `text` inventory. A critical step here is handling the 'l' and 'o' characters. Since each "balloon" requires two 'l's and two 'o's, the available counts for 'l' and 'o' are effectively halved to determine how many *sets* of these letters we possess for forming complete words. For example, if we have 5 'l's, we can only form 2 sets of 'll'. Finally, the solution identifies the minimum value among the available counts for 'b', 'a', 'n', and the *halved* counts for 'l' and 'o'. This minimum value dictates the maximum number of "balloons" we can make, as it represents the character that will be depleted first, acting as our limiting factor.

Maximum Number of Balloons - Complete Solution Guide

Maximum Number of Balloons is LeetCode problem 1189, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string text , you want to use the characters of text to form as many instances of the word "balloon" as possible. You can use each character in text at most once . Return the maximum number of instances that can be formed. Example 1: Input: text = "nlaebolko" Output: 1 Example 2: Input: text = "loonbalxballpoon" Output: 2 Example 3: Input: text = "leetcode" Output: 0 Constraints: 1 <= text.length <= 10 4 text consists of lower case English letters only. Note: This question is the same as

Detailed Explanation

This problem asks us to determine the absolute maximum number of times we can spell out the word "balloon" using characters from a given input string `text`. The crucial constraint is that each character from `text` can only be used once. Think of it like having a limited supply of Scrabble tiles, and you're trying to build the word "balloon" as many times as possible. What makes this particular word interesting is its letter composition: it requires one 'b', one 'a', one 'n', but *two* 'l's and *two* 'o's. This non-uniform requirement for letters 'l' and 'o' is a key aspect to consider. For example, if your `text` is `nlaebolko`, you have exactly enough for one "balloon". But if `text` were `leetcode`, despite having 'l's and 'o's, you simply can't form any "balloons" because crucial letters like 'b' and 'a' are missing entirely. The challenge boils down to managing character inventories and identifying which letter's scarcity will ultimately limit your production.

Solution Approach

The provided solution elegantly solves this by first creating a comprehensive inventory of all characters available in the input `text`. It accomplishes this using a hash map (or dictionary in Python), where each character from `text` becomes a key, and its corresponding value is the count of how many times it appears. This gives us an immediate, efficient lookup for any character's availability. Once this character inventory is built, the solution focuses on the specific characters needed for "balloon": 'b', 'a', 'l', 'o', 'n'. It gathers the counts for these five letters from our `text` inventory. A critical step here is handling the 'l' and 'o' characters. Since each "balloon" requires two 'l's and two 'o's, the available counts for 'l' and 'o' are effectively halved to determine how many *sets* of these letters we possess for forming complete words. For example, if we have 5 'l's, we can only form 2 sets of 'll'. Finally, the solution identifies the minimum value among the available counts for 'b', 'a', 'n', and the *halved* counts for 'l' and 'o'. This minimum value dictates the maximum number of "balloons" we can make, as it represents the character that will be depleted first, acting as our limiting factor.

Step-by-Step Algorithm

Step 1: Initialize a hash map (or array) to store the frequency of each character in the input string. Iterate through the string and increment the count for each character.
Step 2: Check if the counts of 'b', 'a', 'l', 'o', 'n' are all greater than 0. If any of these counts are 0, return 0 as no balloon can be formed.
Step 3: Calculate the number of 'balloons' that can be formed by dividing the count of each character ('l' and 'o' divided by 2) and taking the minimum of these values.
Step 4: Return the minimum count calculated in Step 3 which represents the maximum number of "balloons" formable.

Key Insights

**Efficient Character Frequency Tracking**: The initial step of constructing a hash map (dictionary) for `text` characters is crucial. It provides an O(1) average time complexity lookup for character counts, allowing for rapid assessment of available resources without needing repeated string scans.
**Handling Asymmetrical Letter Requirements**: The word 'balloon' does not require all its letters with the same frequency ('b', 'a', 'n' once; 'l', 'o' twice). The solution correctly accounts for this by dividing the available counts of 'l' and 'o' by two, effectively converting their raw counts into 'pairs' that satisfy the word's specific structure.
**The 'Bottleneck' Principle**: The core idea is that the maximum number of words you can form is always limited by the character that is *most scarce relative to its need*. By taking the `min` across the adjusted counts of all required 'balloon' characters, the solution precisely identifies this bottleneck, ensuring we don't attempt to form more words than our weakest link allows.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Tesla, Wayfair.

Maximum Number of Balloons | LeetCode 1189

Easy

Hash Table String Counting

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

Solution Code

Available in:

class Solution {
    public int maxNumberOfBalloons(String text) {
        int[] count = new int[26];
        for (char c : text.toCharArray()) {
            count[c - 'a']++;
        }
        int min = Integer.MAX_VALUE;
        min = Math.min(min, count['b' - 'a']);
        min = Math.min(min, count['a' - 'a']);
        min = Math.min(min, count['l' - 'a'] / 2);
        min = Math.min(min, count['o' - 'a'] / 2);
        min = Math.min(min, count['n' - 'a']);
        return min;
    }
}

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Problem Info

Description

Problem 1189: Maximum Number of Balloons - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

TeslaWayfair

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1189

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

4 steps

1
Step 1: Initialize a hash map (or array) to store the frequency of each character in the input string. Iterate through the string and increment the count for each character.
2
Step 2: Check if the counts of 'b', 'a', 'l', 'o', 'n' are all greater than 0. If any of these counts are 0, return 0 as no balloon can be formed.
3
Step 3: Calculate the number of 'balloons' that can be formed by dividing the count of each character ('l' and 'o' divided by 2) and taking the minimum of these values.
4
Step 4: Return the minimum count calculated in Step 3 which represents the maximum number of "balloons" formable.

💡

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 character counting loop runs once through the input string. All other operations are constant time. Therefore, time complexity is dominated by this loop. The space used to store character counts is fixed, regardless of the input string's length.

📖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 a single loop for counting and min calculation

Instead of separate loops for counting and finding the minimum, you can combine them into one pass. During each character iteration, update both the counts and simultaneously track the minimum.

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

Trade-offs: This approach might slightly improve readability and reduce the number of loop iterations but only by a constant factor so the time complexity remains O(n).

✨

Key Insights

1**Efficient Character Frequency Tracking**: The initial step of constructing a hash map (dictionary) for `text` characters is crucial. It provides an O(1) average time complexity lookup for character counts, allowing for rapid assessment of available resources without needing repeated string scans.
2**Handling Asymmetrical Letter Requirements**: The word 'balloon' does not require all its letters with the same frequency ('b', 'a', 'n' once; 'l', 'o' twice). The solution correctly accounts for this by dividing the available counts of 'l' and 'o' by two, effectively converting their raw counts into 'pairs' that satisfy the word's specific structure.
3**The 'Bottleneck' Principle**: The core idea is that the maximum number of words you can form is always limited by the character that is *most scarce relative to its need*. By taking the `min` across the adjusted counts of all required 'balloon' characters, the solution precisely identifies this bottleneck, ensuring we don't attempt to form more words than our weakest link allows.

📚

Key Terms Explained

For Students

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 is empty or does not contain all the letters of "balloon". The solution handles this correctly by returning 0.
Edge case 2: The input string contains an abundance of some letters but insufficient quantities of others. The solution correctly determines the maximum number of "balloons" based on the least frequent character (considering 'l' and 'o' are needed twice).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to divide the counts of 'l' and 'o' by 2 because they appear twice in "balloon". This can lead to an incorrect answer.

✗Common mistake 2: Not handling the case where the input string doesn't contain all the required characters. Returning a non-zero answer when a "balloon" can't be formed is incorrect.

✅

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: Inventory management: Determining the maximum number of products that can be assembled given the available components. Each component corresponds to a letter in the "balloon" example and its count is the number of that component in stock.

🏦Real-world scenario 2: Resource allocation: Finding the maximum number of tasks that can be completed given limited resources. Each resource corresponds to a letter and its count represents the quantity of that resource available.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: September 18, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

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.

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Maximum Number of Balloons"?

For this easy-level Hash Table 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.

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