What is the time complexity of Maximum Number of Occurrences of a Substring?

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

What data structures are used to solve Maximum Number of Occurrences of a Substring?

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

What are the key insights for solving Maximum Number of Occurrences of a Substring?

Insight 1: Since we're looking for the *maximum* number of occurrences, we only need to consider substrings of length `minSize`. Substrings longer than `minSize` can be shortened to `minSize` without affecting the number of unique characters or the possibility of finding a matching substring, because any occurrence of a substring with length larger than minSize also implies occurrence of minSize substring within. Insight 2: Use a sliding window of size `minSize` to extract all possible substrings efficiently. Then, count occurrences of only the valid substrings which meet the unique character constraint.

Which companies ask LeetCode 1297. Maximum Number of Occurrences of a Substring in interviews?

This problem is frequently asked at Atlassian, Hubspot, Roblox.

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Maximum Number of Occurrences of a Substring - LeetCode 1297 Solution

Maximum Number of Occurrences of a Substring - Complete Solution Guide

Maximum Number of Occurrences of a Substring is LeetCode problem 1297, 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

Given a string s , return the maximum number of occurrences of any substring under the following rules: The number of unique characters in the substring must be less than or equal to maxLetters . The substring size must be between minSize and maxSize inclusive. Example 1: Input: s = "aababcaab", maxLetters = 2, minSize = 3, maxSize = 4 Output: 2 Explanation: Substring "aab" has 2 occurrences in the original string. It satisfies the conditions, 2 unique letters and size 3 (between minSize and max

Detailed Explanation

The problem asks us to find the maximum number of occurrences of a substring within a given string `s`, subject to two constraints: (1) the substring must contain no more than `maxLetters` unique characters, and (2) its length must be between `minSize` and `maxSize` (inclusive). The function needs to return the highest frequency of any substring meeting these criteria. The string `s` consists of lowercase English letters, and the substrings can overlap.

Solution Approach

The solution uses a sliding window approach to iterate through all possible substrings of length `minSize`. For each substring, it checks if the number of unique characters is less than or equal to `maxLetters`. If it is, the substring's occurrence count is incremented in a hash map. After iterating through all substrings, the solution finds the maximum value among the counts in the hash map, representing the maximum frequency of a valid substring. In essence, this approach finds all possible substring of size `minSize` and check them against the constraints, counting the ones that are valid, and returns the maximal count.

Step-by-Step Algorithm

Step 1: Initialize a hash map (or Counter in Python) called `counts` to store the occurrence count of each valid substring.
Step 2: Iterate through the input string `s` using a sliding window of size `minSize`. The loop variable `i` ranges from 0 to `n - minSize`, where `n` is the length of `s`.
Step 3: For each window, extract the substring `substring = s[i:i + minSize]`.
Step 4: Calculate the number of unique characters in `substring`. This can be done using a `set` in Python or Java, or an equivalent approach in C/C++.
Step 5: Check if the number of unique characters is less than or equal to `maxLetters`. If this condition is met:
Step 6: Increment the count of `substring` in the `counts` hash map. If the substring doesn't already exist as a key, add it with a count of 1. Otherwise, increment the existing count.
Step 7: After iterating through all possible substrings, find the maximum value among the counts in the `counts` hash map.
Step 8: Return the maximum count. If the hash map is empty (no valid substrings were found), return 0.

Key Insights

Insight 1: Since we're looking for the *maximum* number of occurrences, we only need to consider substrings of length `minSize`. Substrings longer than `minSize` can be shortened to `minSize` without affecting the number of unique characters or the possibility of finding a matching substring, because any occurrence of a substring with length larger than minSize also implies occurrence of minSize substring within.
Insight 2: Use a sliding window of size `minSize` to extract all possible substrings efficiently. Then, count occurrences of only the valid substrings which meet the unique character constraint.
Insight 3: Using a hash map (or similar data structure like Python's `Counter`) is crucial for efficiently counting the occurrences of each valid substring.

Complexity Analysis

Time Complexity: O(n*minSize)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Sliding Window.

Companies

Asked at: Atlassian, Hubspot, Roblox.

Maximum Number of Occurrences of a Substring | LeetCode 1297

Medium

Hash Table String Sliding Window

Time: O(n*minSize)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
        Map<String, Integer> counts = new HashMap<>();
        int n = s.length();

        for (int i = 0; i <= n - minSize; i++) {
            String substring = s.substring(i, i + minSize);
            
            long distinctChars = substring.chars().distinct().count();

            if (distinctChars <= maxLetters) {
                counts.put(substring, counts.getOrDefault(substring, 0) + 1);
            }
        }

        int maxFrequency = 0;
        for (int frequency : counts.values()) {
            maxFrequency = Math.max(maxFrequency, frequency);
        }
        
        return maxFrequency;
    }
}

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Problem Info

Description

Problem 1297: Maximum Number of Occurrences of a Substring - Detailed explanation and solution approaches.

Topics

Hash Table String Sliding Window

Companies

AtlassianHubspotRoblox

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1297

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

8 steps

1
Step 1: Initialize a hash map (or Counter in Python) called `counts` to store the occurrence count of each valid substring.
2
Step 2: Iterate through the input string `s` using a sliding window of size `minSize`. The loop variable `i` ranges from 0 to `n - minSize`, where `n` is the length of `s`.
3
Step 3: For each window, extract the substring `substring = s[i:i + minSize]`.
4
Step 4: Calculate the number of unique characters in `substring`. This can be done using a `set` in Python or Java, or an equivalent approach in C/C++.
5
Step 5: Check if the number of unique characters is less than or equal to `maxLetters`. If this condition is met:
6
Step 6: Increment the count of `substring` in the `counts` hash map. If the substring doesn't already exist as a key, add it with a count of 1. Otherwise, increment the existing count.
7
Step 7: After iterating through all possible substrings, find the maximum value among the counts in the `counts` hash map.
8
Step 8: Return the maximum count. If the hash map is empty (no valid substrings were found), return 0.

💡

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*minSize)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sliding window which is O(n-minSize+1), and the unique character check, O(minSize). Thus we have O((n-minSize+1)*minSize). We simplify to O(n*minSize) as minSize <= n. The space complexity can be O(n) as each substring of minSize could be valid, creating that many distinct keys in our counts map.

📖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

Trie-based Approach

Use a Trie data structure to store and count the substrings. Insert each substring of length `minSize` that satisfies the `maxLetters` constraint into the Trie. The count associated with each Trie node would represent the number of times that substring appears. This can improve efficiency if there are a lot of repeating substrings.

Time: O(n * minSize) in the worst case (like the original solution), but potentially better if many substrings are repeated.Space: Potentially O(n), depending on the number of distinct valid substrings. Can be much larger than the original solution if there are a lot of distinct substrings of size `minSize`

Trade-offs: More complex to implement than the hash map approach. Might be beneficial if there are a lot of repeated substrings but could be less space-efficient than the hash map if substrings are very distinct.

✨

Key Insights

1Insight 1: Since we're looking for the *maximum* number of occurrences, we only need to consider substrings of length `minSize`. Substrings longer than `minSize` can be shortened to `minSize` without affecting the number of unique characters or the possibility of finding a matching substring, because any occurrence of a substring with length larger than minSize also implies occurrence of minSize substring within.
2Insight 2: Use a sliding window of size `minSize` to extract all possible substrings efficiently. Then, count occurrences of only the valid substrings which meet the unique character constraint.
3Insight 3: Using a hash map (or similar data structure like Python's `Counter`) is crucial for efficiently counting the occurrences of each valid substring.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: Empty string. The code correctly handles an empty string implicitly because the loops will not execute.
Edge case 2: `maxLetters` is greater than 26. This is permissible according to the constraints and doesn't require special handling, as the code only checks against the actual number of unique letters in a substring.
Edge case 3: No valid substrings are found. The Python code uses `max(counts.values() or [0])` to return 0 when `counts` is empty. The Java and C++ code achieve a similar result by initializing the `maxFrequency` or `maxCount` variable to 0 and returning it if no valid substrings are found. The C code does a similar process, but requires more manual handling. This prevents exceptions or incorrect return values.
Edge case 4: `minSize` is close to string length. The sliding window logic automatically adapts, ensuring no out-of-bounds access. The condition `i <= n - minSize` prevents exceeding the string boundaries.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering overlapping substrings. The sliding window must increment by 1 in each iteration to find all possible occurrences.

✗Common mistake 2: Incorrectly calculating the number of unique characters. Using `len(set(substring))` is a concise and correct way in Python. In other languages, ensure a correct implementation to avoid overcounting or undercounting. The number of unique characters is a core condition.

✗Common mistake 3: Ignoring the `maxSize` constraint. The problem stated `minSize <= substring length <= maxSize`, but the solution only uses `minSize` due to the aforementioned optimization.

✗Common mistake 4: Memory management issues in C. The C code needs to carefully allocate and free memory to avoid memory leaks. The provided C code requires manual memory management and can be prone to errors if not handled correctly, for example, double free in some error conditions.

✅

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: Bioinformatics - Identifying frequently occurring DNA sequences within a genome, subject to constraints on the number of unique nucleotides.

🏦Real-world scenario 2: Network Security - Detecting common byte sequences (signatures) in network traffic that might indicate a malware infection or intrusion attempt, within specific size limits and a limited number of distinct byte values.

🎮Real-world scenario 3: Data Compression - Finding frequently occurring short strings for potential use as dictionary entries in lossless data compression algorithms.

💡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

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

⚠️ 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: July 8, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 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 "Maximum Number of Occurrences of a Substring"?

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?