What is the time complexity of Sum of Beauty of All Substrings?

The optimal solution has a time complexity of O(n^3).

What data structures are used to solve Sum of Beauty of All Substrings?

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

What are the key insights for solving Sum of Beauty of All Substrings?

Insight 1: Substrings can be generated by iterating through all possible start and end indices. Insight 2: For each substring, the frequency of each character needs to be calculated.

Which companies ask LeetCode 1781. Sum of Beauty of All Substrings in interviews?

This problem is frequently asked at Fastenal.

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Sum of Beauty of All Substrings - LeetCode 1781 Solution

Q: How to solve Sum of Beauty of All Substrings?

The provided solution uses a brute-force approach to generate all possible substrings of the input string `s`. For each substring, it calculates the frequency of each character, finds the maximum and minimum frequencies, and computes the beauty of the substring. Finally, it sums the beauty values of all substrings to obtain the final result. The solution iterates through all possible starting positions `i` from 0 to n-1, where n is the length of the string. For each starting position, it iterates through all possible ending positions `j` from `i` to n-1, thereby generating all substrings.

Sum of Beauty of All Substrings - Complete Solution Guide

Sum of Beauty of All Substrings is LeetCode problem 1781, 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

The beauty of a string is the difference in frequencies between the most frequent and least frequent characters. For example, the beauty of "abaacc" is 3 - 1 = 2 . Given a string s , return the sum of beauty of all of its substrings. Example 1: Input: s = "aabcb" Output: 5 Explanation: The substrings with non-zero beauty are ["aab","aabc","aabcb","abcb","bcb"], each with beauty equal to 1. Example 2: Input: s = "aabcbaa" Output: 17 Constraints: 1 <= s.length <= 500 s consists of only lowercase E

Detailed Explanation

The problem requires calculating the sum of the "beauty" of all possible substrings of a given string `s`. The beauty of a string is defined as the difference between the frequency of the most frequent character and the frequency of the least frequent character. The input string `s` consists of lowercase English letters, and its length is between 1 and 500, inclusive. The output should be the sum of the beauty of all substrings of `s`.

Solution Approach

The provided solution uses a brute-force approach to generate all possible substrings of the input string `s`. For each substring, it calculates the frequency of each character, finds the maximum and minimum frequencies, and computes the beauty of the substring. Finally, it sums the beauty values of all substrings to obtain the final result. The solution iterates through all possible starting positions `i` from 0 to n-1, where n is the length of the string. For each starting position, it iterates through all possible ending positions `j` from `i` to n-1, thereby generating all substrings.

Step-by-Step Algorithm

Step 1: Initialize `total_beauty` to 0 to store the sum of beauty values of all substrings.
Step 2: Iterate through all possible starting positions `i` of the substring from 0 to `n-1`.
Step 3: For each starting position `i`, create a frequency array `freq` of size 26 to store the frequency of each lowercase English letter.
Step 4: Iterate through all possible ending positions `j` of the substring from `i` to `n-1`.
Step 5: Increment the frequency of the character `s[j]` in the `freq` array.
Step 6: Initialize `max_f` to 0 and `min_f` to infinity.
Step 7: Iterate through the `freq` array to find the maximum and minimum non-zero frequencies.
Step 8: Calculate the beauty of the current substring as `max_f - min_f`.
Step 9: Add the beauty of the current substring to `total_beauty`.
Step 10: After iterating through all possible substrings, return `total_beauty`.

Key Insights

Insight 1: Substrings can be generated by iterating through all possible start and end indices.
Insight 2: For each substring, the frequency of each character needs to be calculated.
Insight 3: The maximum and minimum frequencies need to be determined for each substring.
Insight 4: Use an array of size 26 to efficiently store and update the character frequencies.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Fastenal.

Sum of Beauty of All Substrings | LeetCode 1781

Medium

Hash Table String Counting

Time: O(n^3)Space: O(1)

Solution Code

Available in:

class Solution {
    public int beautySum(String s) {
        int total_beauty = 0;
        int n = s.length();

        for (int i = 0; i < n; i++) {
            int[] freq = new int[26];
            for (int j = i; j < n; j++) {
                freq[s.charAt(j) - 'a'] += 1;

                int max_f = 0;
                int min_f = Integer.MAX_VALUE;

                for (int k = 0; k < 26; k++) {
                    if (freq[k] > 0) {
                        max_f = Math.max(max_f, freq[k]);
                        min_f = Math.min(min_f, freq[k]);
                    }
                }

                total_beauty += (max_f - min_f);
            }
        }

        return total_beauty;
    }
}

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Problem Info

Description

Problem 1781: Sum of Beauty of All Substrings - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

Fastenal

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1781

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

10 steps

1
Step 1: Initialize `total_beauty` to 0 to store the sum of beauty values of all substrings.
2
Step 2: Iterate through all possible starting positions `i` of the substring from 0 to `n-1`.
3
Step 3: For each starting position `i`, create a frequency array `freq` of size 26 to store the frequency of each lowercase English letter.
4
Step 4: Iterate through all possible ending positions `j` of the substring from `i` to `n-1`.
5
Step 5: Increment the frequency of the character `s[j]` in the `freq` array.
6
Step 6: Initialize `max_f` to 0 and `min_f` to infinity.
7
Step 7: Iterate through the `freq` array to find the maximum and minimum non-zero frequencies.
8
Step 8: Calculate the beauty of the current substring as `max_f - min_f`.
9
Step 9: Add the beauty of the current substring to `total_beauty`.
10
Step 10: After iterating through all possible substrings, return `total_beauty`.

💡

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^3)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^3) because there are two nested loops that iterate through all possible substrings (O(n^2)), and inside the inner loop, there is another loop of length 26 to find the min/max frequencies which is O(1). Effectively O(n^2 * 1) = O(n^2). Furthermore, the solution iterates a constant amount of times (26) to compute min_f and max_f. However the nested loops for min/max frequency counts can be considered O(26), or constant. Thus the nested loops contribute O(n^2) operations to the algorithm. Thus the overall runtime is O(n^2 * 26) which simplifies to O(n^3). The space complexity is O(1) because the frequency array `freq` has a fixed size of 26, regardless of the input string size. No other significant data structures are used.

📖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

Sliding Window Optimization

Instead of recalculating the frequency array for each substring, the frequency array can be maintained and updated as the window slides. This could reduce the time complexity to O(n^2).

Time: O(n^2)Space: O(1)

Trade-offs: This approach has a better time complexity, but requires more careful implementation to manage the sliding window and frequency updates. The space complexity remains the same as the original solution.

✨

Key Insights

1Insight 1: Substrings can be generated by iterating through all possible start and end indices.
2Insight 2: For each substring, the frequency of each character needs to be calculated.
3Insight 3: The maximum and minimum frequencies need to be determined for each substring.
4Insight 4: Use an array of size 26 to efficiently store and update the character frequencies.

📚

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: Empty string: The code implicitly handles an empty string because the outer loops won't execute when n=0, correctly returning 0.
Edge case 2: String with only one character: The inner loop will execute only once, max_f and min_f will be the same, resulting in a beauty of 0 for that substring, which is the correct behavior.
Edge case 3: String with all same characters: The beauty of any substring is 0 because max and min frequencies are the same.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize `min_f` to `float('inf')` or `Integer.MAX_VALUE` before finding the minimum frequency. This will result in incorrect minimum frequency calculations.

✗Common mistake 2: Not handling edge cases where the frequency of all characters is zero. The code correctly handles it by skipping in min/max calculation loops but some implementations may have issues.

✅

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: Text analysis: Finding patterns in text where the relative frequency of words or characters is important.

🏦Real-world scenario 2: Data compression: Identifying frequent and infrequent symbols to optimize encoding schemes.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: November 9, 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

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.

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 "Sum of Beauty of All Substrings"?

For this medium-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.

🤖 Need help with this problem?