Maximum Frequency Stack - LeetCode solutions

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]

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

class FreqStack {
    private static class Node {
        Node next;
        int val;

        Node(int val) {
            this.val = val;
            this.next = null;
        }

        Node() {
            this.next = null;
        }
    }

    private static class DLL {
        DLL next;
        int val;
        Node head;
        int size;

        public DLL() {
            head = new Node();
            this.size = 0;
        }

        public void addNode(int x) {
            Node node = new Node(x);
            node.next = head.next;
            head.next = node;
            size++;
        }

        public Node removeNode() {
            Node node = head.next;
            if (node != null) {
                head.next = node.next;
                node.next = null;
                size--;
            }
            return node;
        }
    }

    private int max;
    private HashMap<Integer, Integer> freqMap;
    private HashMap<Integer, DLL> freqListMap;

    public FreqStack() {
        max = 0;
        freqMap = new HashMap<>();
        freqListMap = new HashMap<>();
    }

    public void push(int val) {
        int count = freqMap.getOrDefault(val, 0) + 1;
        max = Math.max(max, count);
        freqMap.put(val, count);
        DLL dll = freqListMap.getOrDefault(count, new DLL());
        dll.addNode(val);
        freqListMap.put(count, dll);
    }

    public int pop() {
        DLL dll = freqListMap.get(max);
        Node node = dll.removeNode();
        freqMap.put(node.val, max - 1);
        if (dll.size == 0) {
            max--;
        }
        return node.val;
    }
}

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

Explain:

nope.

Complexity:

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