Maximum Frequency Stack

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", "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].

Implementation

Stack of Stacks + maintain frequency of every number + counter for max_freq

• Map freq to stack of elements. Update max_freq and frequency of every number accordingly during push/pop

class FreqStack:
 
    def __init__(self):
        self.freq = Counter()
        self.group = defaultdict(list)
        self.maxfreq = 0
        
    def push(self, val: int) -> None:
        f = self.freq[val] + 1
        self.freq[val] = f
        if f > self.maxfreq:
            self.maxfreq = f
        self.group[f].append(val)
        
 
    def pop(self) -> int:
        x = self.group[self.maxfreq].pop()
        self.freq[x] -= 1
        if not self.group[self.maxfreq]:
            self.maxfreq -= 1
        return x

Optimizations

Optimized Complexity

• Time Complexity:  O(1)O(1)O(1) for both push and pop operations.

• Space Complexity:  O(N)O(N)O(N), where N is the number of elements in the FreqStack.

Related

lc-hard: https://leetcode.com/problems/maximum-frequency-stack/