Maximum Frequency Stack

Maximum Frequency Stack

1. What is this problem about?

Maximum Frequency Stack is a design problem where you need to implement a stack-like data structure. However, instead of popping the most recently added element, you pop the most frequent element. If there's a tie for the most frequent element, you pop the one that was added most recently (standard stack behavior).

2. Why is this asked in interviews?

This is a classic 'Design' interview question asked by almost every major tech company (Apple, Uber, Microsoft, Amazon, etc.). It tests your ability to choose and combine the right data structures to achieve O(1) time complexity for both push and pop operations. It evaluates your understanding of Hash Maps and Stacks.

3. Algorithmic pattern used

The optimal solution uses a 'Map of Stacks'.

1. freq: A Hash Map to keep track of the current frequency of each element.
2. group: A Hash Map where the key is a frequency f and the value is a Stack of elements that have a frequency of at least f.
3. maxFreq: An integer to track the current maximum frequency in the stack. When you push an element, you increment its frequency and add it to the stack corresponding to its new frequency in the group map.

4. Example explanation

• Push 5: Freq(5)=1. Group[1] = [5]. maxFreq=1.
• Push 7: Freq(7)=1. Group[1] = [5, 7]. maxFreq=1.
• Push 5: Freq(5)=2. Group[2] = [5]. maxFreq=2.
• Pop: maxFreq is 2. Pop from Group[2] -> returns 5. Freq(5) becomes 1. maxFreq remains 2 (or moves to 1 if Group[2] was empty).
• Pop: maxFreq is 1. Pop from Group[1] -> returns 7 (the most recent).

5. Common mistakes candidates make

A common mistake is using a Priority Queue (Heap). While a heap works, it results in O(log n) complexity. Top-tier companies expect the O(1) 'Map of Stacks' solution. Another mistake is forgetting to handle the 'most recent' tie-breaking rule correctly.

6. Interview preparation tip

When asked to design a data structure with specific 'pop' or 'search' requirements, think about how you can group the data. In this case, grouping elements by their frequencies allows for instant access to the most frequent ones.