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LeetCode: 895-Maximum Frequency Stack 解題紀錄

Last Updated on 2022-03-19 by Clay

題目

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.

題目要我們設計一個類似 stack 的結構，不過若要彈出（pop），則必須要按照數值的頻率來排優先度；若是頻率一樣的數值，則像 stack 一樣越接近頂部越優先彈出。

解題思路

Hash Map 解法

老實說這題還滿容易就 TLE 了，所以比較簡單的做法應該是使用 hash map 來紀錄一個數值的出現次數、以及特定頻率下數值推入的順序。

如果彈出了數值，則需要把該數值對應的出現次數減去一、並將對應最高頻率的順序表彈出。

並且還需要一個紀錄最大出現次數的變數，並在彈出後檢查最大頻率的順序表是否為空；最大頻率對應的順序表若為空，則將最大頻率的變數減去一。

詳細的製作方法請參考程式碼。


C++ 範例程式碼

class FreqStack {
public:
    // Init
    int maxFreq = 0;
    unordered_map<int, int> val2freq;
    unordered_map<int, stack<int>> freq2stack;
    
    FreqStack() {
        
    }
    
    // Push09sxaz
    void push(int val) {
        ++val2freq[val];
        maxFreq = max(maxFreq, val2freq[val]);
        
        freq2stack[val2freq[val]].push(val);
    }
    
    int pop() {
        // Get the top value
        int top = freq2stack[maxFreq].top();
        freq2stack[maxFreq].pop();
        
        // Decrement the frquency of the top value (for the next push)
        --val2freq[top];
        
        // If the maxFreq stack has no any value
        if (freq2stack[maxFreq].empty()) --maxFreq;
        
        return top;
    }
};

/**
 * Your FreqStack object will be instantiated and called as such:
 * FreqStack* obj = new FreqStack();
 * obj->push(val);
 * int param_2 = obj->pop();
 */



兩節點 Python 範例程式碼

class FreqStack:
    def __init__(self):
        self.max_freq = 0
        self.val2freq = dict()
        self.freq2stack = dict()

    def push(self, val: int) -> None:
        freq = self.val2freq.get(val, 0) + 1
        
        self.val2freq[val] = freq
        self.max_freq = max(self.max_freq, freq)
        self.freq2stack[freq] = self.freq2stack.get(freq, []) + [val]

    def pop(self) -> int:
        top = self.freq2stack[self.max_freq].pop()
        self.val2freq[top] -= 1
        
        if len(self.freq2stack[self.max_freq]) == 0:
            self.max_freq -= 1
        
        return top
        


# Your FreqStack object will be instantiated and called as such:
# obj = FreqStack()
# obj.push(val)
# param_2 = obj.pop()

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