Maximum Frequency Stack

Background Knowledge

The "Maximum Frequency Stack" problem involves designing a custom stack-like data structure that supports two primary operations: push(val) and pop(). The key aspect of this problem is that the pop() operation should remove the most frequent element from the stack. If there's a tie in frequency, the element closest to the top of the stack should be removed. This problem falls under the category of Hash Maps & Counting, indicating that we'll likely need to utilize a hash map (or dictionary) to keep track of element frequencies.

To tackle this problem, it's essential to understand the basics of stacks and hash maps. A stack is a linear data structure that follows the Last-In-First-Out (LIFO) principle, where elements are added and removed from the top. A hash map, on the other hand, is a data structure that stores key-value pairs, allowing for efficient lookups, insertions, and deletions. We'll need to leverage these data structures to keep track of element frequencies and their order in the stack.

The problem also requires an understanding of how to handle ties in frequency. This means we'll need to consider the order of elements in the stack and ensure that the most recent element with the highest frequency is removed first. This can be achieved by using additional data structures, such as lists or arrays, to store elements with the same frequency.

Algorithm/Approach

The general approach to solving this problem involves using a combination of a hash map to store element frequencies and a stack (or multiple stacks) to keep track of the order of elements. We'll need to design an algorithm that updates the frequency count of each element when push(val) is called and removes the most frequent element (or the most recent one in case of a tie) when pop() is called.

Step-by-Step Strategy

To implement the solution, follow these steps:

• Create a hash map to store the frequency of each element.
• Design a data structure to store the order of elements (e.g., a stack or list).
• When push(val) is called:
  • Increment the frequency count of the element in the hash map.
  • Add the element to the order data structure.
• When pop() is called:
  • Find the most frequent element(s) in the hash map.
  • If there's a tie, identify the most recent element(s) in the order data structure.
  • Remove the most frequent (or most recent) element from the order data structure and update the frequency count in the hash map.
• Repeat the pop() process until the desired result is achieved.

Common Pitfalls

When implementing the solution, watch out for the following common pitfalls:

• Failing to update the frequency count correctly when push(val) is called.
• Not handling ties in frequency correctly, leading to incorrect pop() results.
• Using an inefficient data structure to store the order of elements, resulting in poor performance.

Time & Space Complexity

The expected time complexity for the push(val) operation is $O(1)$, as we only need to update the frequency count and add the element to the order data structure. The time complexity for the pop() operation is $O(n)$ in the worst case, where $n$ is the number of elements in the stack, as we need to find the most frequent element(s) and update the frequency count. The space complexity is $O(n)$, as we need to store the frequency count and order of all elements.