Maximum Frequency Stack

Implement FreqStack, a class which simulates the operation of a stack-like data structure.

FreqStack has two functions:

  • push(int x), which pushes an integer x onto the stack.

  • pop(), which removes and returns the most frequent element in the stack.

    • If there is a tie for most frequent element, the element closest to the top of the stack 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:
After making six .push operations, the stack is [5,7,5,7,4,5] from bottom to top.  Then:

pop() -> returns 5, as 5 is the most frequent.
The stack becomes [5,7,5,7,4].

pop() -> returns 7, as 5 and 7 is the most frequent, but 7 is closest to the top.
The stack becomes [5,7,5,4].

pop() -> returns 5.
The stack becomes [5,7,4].

pop() -> returns 4.
The stack becomes [5,7].

Note:

  • Calls to FreqStack.push(int x) will be such that 0 <= x <= 10^9.

  • It is guaranteed that FreqStack.pop() won't be called if the stack has zero elements.

  • The total number of FreqStack.push calls will not exceed 10000 in a single test case.

  • The total number of FreqStack.pop calls will not exceed 10000 in a single test case.

  • The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.

class FreqStack {
    // Map of element -> frequency
    HashMap<Integer, Integer> freq;
    // Map of frequency -> stack of elements with this frequency
    HashMap<Integer, Stack<Integer>> m;
    int maxFreq;

    public FreqStack() {
        freq = new HashMap<>();
        m = new HashMap<>();
        maxFreq = 0;
    }

    public void push(int x) {
        int f = freq.getOrDefault(x, 0) + 1;
        freq.put(x, f);
        maxFreq = Math.max(maxFreq, f);
        if (!m.containsKey(f))
            m.put(f, new Stack<Integer>());
        m.get(f).add(x);
    }

    // Remember that map of all the frequencies upto the maxFreq exists
    // Because when we promote an element from freq to freq+1
    // then we are not actually removing that element from that stack
    public int pop() {
        int x = m.get(maxFreq).pop();
        freq.put(x, maxFreq - 1);
        if (m.get(maxFreq).size() == 0)
            maxFreq--;
        return x;
    }
}

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