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895. Maximum Frequency Stack ​

题目 ​

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.
Related Topics
  • 栈
  • 设计
  • 哈希表
  • 有序集合

  • 👍 433
  • 👎 0

    • 思路 ​

    题目中模拟 frequency stack 时，直观模拟一维数组结构，

    内部实现如果采用单一维数组作为数据结构，每次查找处理元素需要对其 frequency 进行排序， 则每次查找效率为 O(nlogn)

    对于 Frequency Stack 这种数据模式，我们只需要关心 push 和 pop 操作的返回值，无需关心所有数据的排序

    需要优化查找复杂性, 可根据词频进行数据存储, 每次 pop 查找时返回词频最大的数组中的首个元素即可。

    注意，词频统计是连续的，不存在跳跃的情况

    • 当前最大词频是 8，
    • 词频为 8 的元素都被 pop 后， 即 len(self.group[8]) == 0 or not self.group[8]
    • 则下一个最大词频则是 7

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
from collections import defaultdict
from itertools import count
from typing import List

class FreqStack:

    def __init__(self):
        self.group = defaultdict(list)
        self.count = defaultdict(int)
        self.max_freq = 0

    def push(self, val: int) -> None:
        freq = self.count[val] + 1
        self.group[freq].append(val)
        self.count[val] = freq

        self.max_freq = max(self.max_freq, freq)


    def pop(self) -> int:

        max_freq_group =  self.group[self.max_freq]
        val = max_freq_group.pop()
        self.count[val] -= 1

        if not self.group[self.max_freq]:
            self.max_freq -= 1

        return val




# Your FreqStack object will be instantiated and called as such:
# obj = FreqStack()
# obj.push(val)
# param_2 = obj.pop()
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(1)
    • 空间复杂度 O(N)