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有趣的一題

Problem

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]ExplanationFreqStack 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.

Sol

class FreqStack:

    def __init__(self):
        self.stks = defaultdict(list)
        self.v2l = defaultdict(int)
        self.max_len = 0

    def push(self, val: int) -> None:
        self.v2l[val] += 1
        self.stks[self.v2l[val]].append(val)
        self.max_len = max(self.max_len, self.v2l[val])

    def pop(self) -> int:
        val = self.stks[self.max_len].pop()
        if not self.stks[self.max_len]:
            self.max_len -= 1
        
        self.v2l[val] -= 1
        return val