[LeetCode February Challange] Day 28 - Maximum Frequency Stack

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 <= 10^9
• At most 2 * 10^4 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.

---

Solution

Time complexity : O(1)
Space complexity : O(n)

class FreqStack {
public:
    FreqStack() {
        maxFreq = 0;
        freqM = {};
        freqRecentM = {};
    }
    
    void push(int val) {
        if (maxFreq < ++freqM[val])
            maxFreq = freqM[val];
        freqRecentM[freqM[val]].push(val);
    }
    
    int pop() {
        int res = freqRecentM[maxFreq].top();
        freqRecentM[maxFreq].pop();
        --freqM[res];
        if (freqRecentM[maxFreq].empty())
            --maxFreq;
        return res;
    }
private:
    int maxFreq;
    unordered_map<int, int> freqM;
    unordered_map<int, stack<int>> freqRecentM;
};

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

當 push 時：

1. 更新該數的 freq 值。
2. 更新 maxFreq 值。
3. 在該 freq 出現的順序 stack 中加入該數。

當 pop 時：

1. 取出目前 maxFreq 順序中最新的那一個。
2. 該數的 freq -1。
3. 若 maxFreq 的 stack 為空，則 maxFreq 往下減少 1。