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].

 

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.

Ideas:

1. 利用count 来记录每个num 出现的次数总和， 并且一直update

2. 利用self.maxFreq 来记录当前num出现最对的次数，可能有tie

3. 利用self.freqs 来记录每个次数的nums, 并且用stack来记录先后顺序

4. 每次pop 之后要检查self.freqs[self.maxFreq] 是否为空，然后来update self.maxFreq.

 

Code:

class FreqStack:

    def __init__(self):
        self.count = collections.Counter()
        self.freqs = collections.defaultdict(list)
        self.maxFreq = 0
        

    def push(self, val: int) -> None:
        self.count[val] += 1
        val_count = self.count[val]
        self.freqs[val_count].append(val)
        self.maxFreq = max(self.maxFreq, val_count)
        

    def pop(self) -> int:
        val = self.freqs[self.maxFreq].pop()
        self.count[val] -= 1
        if not self.freqs[self.maxFreq]:
            self.maxFreq -= 1
        return val