How do you solve Maximum Frequency Stack on LeetCode?

Maximum Frequency Stack (LeetCode #895) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a frequency map allows efficient tracking of element counts.

What is the brute force solution for Maximum Frequency Stack?

The brute force approach for Maximum Frequency Stack is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we maintain a list of elements and their frequencies. Each time we pop, we scan through the list to find the most frequent element. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Frequency Stack?

Maximum Frequency Stack uses the following concepts: Hash Table, Stack, Design, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Frequency Stack?

Clearly explain your thought process while coding. Consider edge cases, such as popping from an empty stack. Practice implementing both brute force and optimal solutions.

What are common mistakes when solving Maximum Frequency Stack?

Not updating the max frequency correctly after pops. Confusing the frequency of elements with their order in the stack.

#895

Maximum Frequency Stack

Hard

Hash TableStackDesignOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

In the optimal approach, we use a combination of a frequency map and a stack of stacks. Each frequency level has its own stack, allowing us to efficiently manage and pop the most frequent elements.

⚙️

Algorithm

4 steps

1Step 1: Use a frequency map to count occurrences of each element.
2Step 2: Use a map of stacks to store elements by their frequency.
3Step 3: For each push, increment the frequency and push the element onto the corresponding stack.
4Step 4: For each pop, retrieve the stack of the highest frequency and pop the top element.

solution.py22 lines

1# Full working Python code
2class FreqStack:
3    def __init__(self):
4        self.freq = {}
5        self.freq_stack = {}
6        self.max_freq = 0
7
8    def push(self, val: int) -> None:
9        self.freq[val] = self.freq.get(val, 0) + 1
10        f = self.freq[val]
11        if f > self.max_freq:
12            self.max_freq = f
13        if f not in self.freq_stack:
14            self.freq_stack[f] = []
15        self.freq_stack[f].append(val)
16
17    def pop(self) -> int:
18        val = self.freq_stack[self.max_freq].pop()
19        self.freq[val] -= 1
20        if not self.freq_stack[self.max_freq]:
21            self.max_freq -= 1
22        return val

ℹ

Complexity note: The time complexity is O(n) for push and pop operations because we are using a hashmap and stacks to manage frequencies and elements efficiently, allowing us to access and modify them in constant time.

1Using a frequency map allows efficient tracking of element counts.
2Stack of stacks enables O(1) access to the most frequent elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.