What is the time complexity of Maximum Frequency Stack?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Maximum Frequency Stack?

This problem uses Hash Table, Stack, Design, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Frequency Stack?

Insight 1: A hashmap (or dictionary) is needed to efficiently track the frequency of each element. Insight 2: Another hashmap is needed to store elements grouped by their frequencies, with each frequency mapping to a stack of elements having that frequency.

Which companies ask LeetCode 895. Maximum Frequency Stack in interviews?

This problem is frequently asked at Nutanix, PayPal, Salesforce, Snap.

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Maximum Frequency Stack - LeetCode 895 Solution

Maximum Frequency Stack - Complete Solution Guide

Maximum Frequency Stack is LeetCode problem 895, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack. Implement the FreqStack class: FreqStack() constructs an empty frequency stack. void push(int val) pushes an integer val onto the top of the stack. int pop() removes and returns the most frequent element in the stack. If there is a tie for the most frequent element, the element closest to the stack's top is removed and returned. Example 1: Input ["FreqStack", "push", "push", "push",

Detailed Explanation

The problem requires designing a stack-like data structure called `FreqStack`. This data structure needs to support two operations: `push(val)` which adds an integer `val` to the top of the stack, and `pop()` which removes and returns the most frequent element in the stack. If multiple elements have the same highest frequency, the element closest to the top of the stack is returned.

Solution Approach

The solution uses two hashmaps and a `max_freq` variable. The `freq` hashmap stores the frequency of each element. The `group` hashmap stores a list of elements for each frequency level. When an element is pushed, its frequency is updated, and it is added to the corresponding list in `group`. The `max_freq` variable is updated whenever a new maximum frequency is encountered. When popping, the element from the top of the list associated with `max_freq` is removed, its frequency is decremented, and `max_freq` is decremented if the list becomes empty.

Step-by-Step Algorithm

Step 1: Initialize `freq` (frequency hashmap), `group` (frequency-based stack hashmap), and `max_freq` to 0.
Step 2: In `push(val)`: Increment `freq[val]`.
Step 3: Update `max_freq` if `freq[val]` is greater than the current `max_freq`.
Step 4: Append `val` to `group[freq[val]]` (the stack associated with its new frequency).
Step 5: In `pop()`: Get the element at the top of `group[max_freq]` and remove it.
Step 6: Decrement the frequency of the popped element in `freq`.
Step 7: If `group[max_freq]` is now empty, decrement `max_freq`.
Step 8: Return the popped element.

Key Insights

Insight 1: A hashmap (or dictionary) is needed to efficiently track the frequency of each element.
Insight 2: Another hashmap is needed to store elements grouped by their frequencies, with each frequency mapping to a stack of elements having that frequency.
Insight 3: Maintain a `max_freq` variable to keep track of the highest frequency seen so far. This allows O(1) access to the stack containing elements with the maximum frequency.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(N)

Topics

This problem involves: Hash Table, Stack, Design, Ordered Set.

Companies

Asked at: Nutanix, PayPal, Salesforce, Snap.

Maximum Frequency Stack | LeetCode 895

Hard

Hash Table Stack Design Ordered Set

Time: O(1)Space: O(N)

Solution Code

Available in:

import java.util.HashMap;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;

class FreqStack {

    Map<Integer, Integer> freq;
    Map<Integer, List<Integer>> group;
    int max_freq;

    public FreqStack() {
        freq = new HashMap<>();
        group = new HashMap<>();
        max_freq = 0;
    }

    public void push(int val) {
        freq.put(val, freq.getOrDefault(val, 0) + 1);
        int f = freq.get(val);
        if (f > max_freq) {
            max_freq = f;
        }
        group.computeIfAbsent(f, k -> new ArrayList<>()).add(val);
    }

    public int pop() {
        List<Integer> list = group.get(max_freq);
        int val = list.remove(list.size() - 1);
        freq.put(val, freq.get(val) - 1);
        if (list.isEmpty()) {
            max_freq--;
        }
        return val;
    }
}

/**
 * Your FreqStack object will be instantiated and called as such:
 * FreqStack obj = new FreqStack();
 * obj.push(val);
 * int param_2 = obj.pop();
 */

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Problem Info

Description

Problem 895: Maximum Frequency Stack - Detailed explanation and solution approaches.

Topics

Hash Table Stack Design Ordered Set

Companies

NutanixPayPalSalesforceSnap

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:895

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 `freq` (frequency hashmap), `group` (frequency-based stack hashmap), and `max_freq` to 0.
2
Step 2: In `push(val)`: Increment `freq[val]`.
3
Step 3: Update `max_freq` if `freq[val]` is greater than the current `max_freq`.
4
Step 4: Append `val` to `group[freq[val]]` (the stack associated with its new frequency).
5
Step 5: In `pop()`: Get the element at the top of `group[max_freq]` and remove it.
6
Step 6: Decrement the frequency of the popped element in `freq`.
7
Step 7: If `group[max_freq]` is now empty, decrement `max_freq`.
8
Step 8: Return the popped element.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(N)

📝

Why?

The `push` and `pop` operations involve hashmap lookups and appends/removals from lists, all of which are O(1) on average. The space complexity is O(N) because, in the worst case, we might store N distinct elements in the `freq` and `group` hashmaps, where N is the total number of `push` operations.

📖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

Priority Queue with Timestamp

Use a priority queue to store the elements based on their frequency and a timestamp. The timestamp is used to break ties among elements with the same frequency. This simulates the stack-like behavior.

Time: O(log N) for push and pop operations due to priority queue operations.Space: O(N)

Trade-offs: This approach has logarithmic time complexity for `push` and `pop` compared to O(1) of the main solution, but provides an alternative if the underlying data structure does not provide amortized O(1) for hashmap operations. Might be more robust to pathological cases for certain hashmap implementations.

✨

Key Insights

1Insight 1: A hashmap (or dictionary) is needed to efficiently track the frequency of each element.
2Insight 2: Another hashmap is needed to store elements grouped by their frequencies, with each frequency mapping to a stack of elements having that frequency.
3Insight 3: Maintain a `max_freq` variable to keep track of the highest frequency seen so far. This allows O(1) access to the stack containing elements with the maximum frequency.

📚

Key Terms Explained

For Students

Hash Table

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

Stack

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

Edge Cases

Edge case 1: All elements pushed are the same. The frequency of that element will continuously increase and the `max_freq` will also increase linearly. The group dictionary will contain stacks for frequencies 1, 2, 3, ..., `max_freq`. The solution correctly handles this.
Edge case 2: The stack is empty when `pop` is called (violating the constraint). While the problem statement guarantees that the stack will not be empty, robust implementations can add an additional check for stack emptiness.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to decrement `max_freq` when the list corresponding to the maximum frequency becomes empty. This leads to incorrect `pop` operations.

✗Common mistake 2: Incorrectly updating the frequency of the popped element, potentially leading to incorrect future pops.

✅

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: Cache eviction policies where frequently accessed items are prioritized. The `FreqStack` can be adapted to track access frequency and evict less frequent items.

🏦Real-world scenario 2: Text editors with an 'undo' feature. The most recent and frequent actions are often the ones a user wants to 'undo', making this approach beneficial.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 23, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 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 a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Frequency Stack"?

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

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