#895

Hard

auto_awesomeStack-based state management

LeetCode Problem Workspace

Maximum Frequency Stack

Design a stack-like data structure to manage elements and handle frequent stack operations, including popping the most frequent element.

hash table stack design ordered set

Problem Statement

Given a stack-like data structure, design an efficient solution for pushing elements to the stack and popping the most frequent element. Each element can appear multiple times, and when popped, the most frequent element should be removed first. If multiple elements have the same frequency, the element closest to the top of the stack should be popped.

You need to implement the FreqStack class, which will support the push and pop operations efficiently. The operations should be optimized for performance considering the constraints that you may need to handle large numbers of push and pop calls.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["FreqStack", "push", "push", "push", "push", "push", "push", "pop", "pop", "pop", "pop"] [[], [5], [7], [5], [7], [4], [5], [], [], [], []] Output [null, null, null, null, null, null, null, 5, 7, 5, 4]

Explanation FreqStack freqStack = new FreqStack(); freqStack.push(5); // The stack is [5] freqStack.push(7); // The stack is [5,7] freqStack.push(5); // The stack is [5,7,5] freqStack.push(7); // The stack is [5,7,5,7] freqStack.push(4); // The stack is [5,7,5,7,4] freqStack.push(5); // The stack is [5,7,5,7,4,5] freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4]. freqStack.pop(); // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4]. freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,4]. freqStack.pop(); // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].

Constraints

0 <= val <= 109
At most 2 * 104 calls will be made to push and pop.
It is guaranteed that there will be at least one element in the stack before calling pop.

Solution Approach

Hash Table for Frequency Tracking

Use a hash table to track the frequency of each element. Each element will map to its frequency count, enabling efficient retrieval of the most frequent element.

Stack for Order Maintenance

Use a stack to store elements in a way that maintains the order of their insertion. When popping the most frequent element, the stack's order ensures that the closest element is popped first.

Max Frequency Management

Use a secondary hash table or a priority queue to maintain and quickly access the most frequent element in the stack. This approach allows for efficient popping based on frequency while adhering to the stack's LIFO behavior.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of push and pop operations depends on the final implementation. In the optimal approach, the frequency tracking and stack management should both allow for O(1) push and pop operations. However, managing the frequency order can introduce extra overhead depending on the approach chosen for max frequency tracking, leading to O(log n) or O(1) depending on implementation.

What Interviewers Usually Probe

Assess how well the candidate manages stack-based state manipulation and tracking element frequencies.
Evaluate the candidate's ability to optimize the problem considering both time and space constraints.
Test if the candidate can balance simplicity with efficiency, especially when managing large numbers of operations.

Common Pitfalls or Variants

Common pitfalls

Failing to manage the frequency of elements efficiently can lead to suboptimal performance.
Not correctly handling edge cases where multiple elements have the same frequency.
Overcomplicating the solution by using overly complex data structures when a simpler one will suffice.

Follow-up variants

Allowing multiple elements with the same frequency and managing the frequency stack differently.
Implementing a solution that supports different stack management strategies, like using a priority queue.
Optimizing for a different performance trade-off, such as focusing more on time complexity over space.

FAQ

How do I solve the Maximum Frequency Stack problem?

Solve this problem by using a combination of hash tables to track frequencies and stacks to maintain element order. Optimize for both time and space.

What is the main challenge in the Maximum Frequency Stack problem?

The challenge lies in efficiently managing the stack while considering element frequencies and ensuring the most frequent element is popped first.

How does the stack-based state management apply to this problem?

In this problem, stack-based state management ensures that the order of element insertion is preserved, while frequency-based popping ensures the most frequent element is prioritized.

Can I use a priority queue for the Maximum Frequency Stack problem?

Yes, a priority queue could be used to manage the maximum frequency efficiently. However, managing the stack order might require additional considerations.

How does GhostInterview help with Maximum Frequency Stack?

GhostInterview helps by providing targeted strategies for managing frequency and stack-based operations, optimizing both time and space complexity for large input sizes.

terminal

Solution

Solution 1: Hash Table + Priority Queue (Max Heap)

According to the problem description, we need to design a data structure that supports popping out the element with the highest frequency. If multiple elements have the same frequency, the element closest to the top of the stack should be popped out.

class FreqStack:
    def __init__(self):
        self.cnt = defaultdict(int)
        self.q = []
        self.ts = 0

    def push(self, val: int) -> None:
        self.ts += 1
        self.cnt[val] += 1
        heappush(self.q, (-self.cnt[val], -self.ts, val))

    def pop(self) -> int:
        val = heappop(self.q)[2]
        self.cnt[val] -= 1
        return val


# Your FreqStack object will be instantiated and called as such:
# obj = FreqStack()
# obj.push(val)
# param_2 = obj.pop()

Solution 2: Double Hash Tables

In Solution 1, in order to pop out the required element, we maintained a priority queue and had to operate on it each time, which has a time complexity of $O(\log n)$. If we can find the required element in $O(1)$ time, then the time complexity of each operation of the entire data structure can be reduced to $O(1)$.

class FreqStack:
    def __init__(self):
        self.cnt = defaultdict(int)
        self.q = []
        self.ts = 0

    def push(self, val: int) -> None:
        self.ts += 1
        self.cnt[val] += 1
        heappush(self.q, (-self.cnt[val], -self.ts, val))

    def pop(self) -> int:
        val = heappop(self.q)[2]
        self.cnt[val] -= 1
        return val


# Your FreqStack object will be instantiated and called as such:
# obj = FreqStack()
# obj.push(val)
# param_2 = obj.pop()

Continue Practicing

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