Maximum Frequency Stack Python Solution

lcpy gen -n 895   # by problem number
lcpy gen -s maximum_frequency_stack   # by problem name

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]

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].

class FreqStack:
    def __init__(self) -> None:
        self.freq: dict[int, int] = {}
        self.group: dict[int, list[int]] = {}
        self.max_freq = 0

    # Time: O(1)
    # Space: O(n)
    def push(self, val: int) -> None:
        count = self.freq.get(val, 0) + 1
        self.freq[val] = count
        if count > self.max_freq:
            self.max_freq = count
        self.group.setdefault(count, []).append(val)

    # Time: O(1)
    # Space: O(n)
    def pop(self) -> int:
        val = self.group[self.max_freq].pop()
        self.freq[val] -= 1
        if not self.group[self.max_freq]:
            self.max_freq -= 1
        return val