895. Maximum Frequency Stack - Explanation

Description

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 <= 1,000,000,000
At most 20,000 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.

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Stacks - The core data structure used to maintain elements by frequency level
Hash Maps - Used for tracking element frequencies and grouping elements by frequency
Heaps/Priority Queues - One solution approach uses heaps to efficiently retrieve the most frequent element
Data Structure Design - Understanding how to combine multiple data structures to achieve efficient operations

1. Brute Force

Intuition

The most straightforward approach is to maintain a regular stack along with a frequency count for each element. When we need to pop, we find the maximum frequency among all elements, then scan backwards through the stack to find the most recent element with that frequency. While simple to understand, this requires scanning the entire stack on every pop operation.

Algorithm

Initialize a hash map cnt to track the frequency of each element and a list stack to store pushed elements.
For push(val): append the value to the stack and increment its count in the hash map.
For pop(): find the maximum frequency across all elements, then iterate from the end of the stack to find the first element with that frequency. Remove it from the stack, decrement its count, and return it.

class FreqStack:

    def __init__(self):
        self.cnt = defaultdict(int)
        self.stack = []

    def push(self, val: int) -> None:
        self.stack.append(val)
        self.cnt[val] += 1

    def pop(self) -> int:
        maxCnt = max(self.cnt.values())
        i = len(self.stack) - 1
        while self.cnt[self.stack[i]] != maxCnt:
            i -= 1
        self.cnt[self.stack[i]] -= 1
        return self.stack.pop(i)

public class FreqStack {
    private Map<Integer, Integer> cnt;
    private List<Integer> stack;

    public FreqStack() {
        cnt = new HashMap<>();
        stack = new ArrayList<>();
    }

    public void push(int val) {
        stack.add(val);
        cnt.put(val, cnt.getOrDefault(val, 0) + 1);
    }

    public int pop() {
        int maxCnt = Collections.max(cnt.values());
        int i = stack.size() - 1;
        while (cnt.get(stack.get(i)) != maxCnt) {
            i--;
        }
        int val = stack.remove(i);
        cnt.put(val, cnt.get(val) - 1);
        return val;
    }
}

class FreqStack {
private:
    unordered_map<int, int> cnt;
    vector<int> stack;

public:
    FreqStack() {}

    void push(int val) {
        stack.push_back(val);
        cnt[val]++;
    }

    int pop() {
        int maxCnt = 0;
        for (auto& [_, frequency] : cnt) {
            maxCnt = max(maxCnt, frequency);
        }
        int i = stack.size() - 1;
        while (cnt[stack[i]] != maxCnt) {
            i--;
        }
        int val = stack[i];
        stack.erase(stack.begin() + i);
        cnt[val]--;
        return val;
    }
};

class FreqStack {
    constructor() {
        this.cnt = new Map();
        this.stack = [];
    }

    /**
     * @param {number} val
     * @return {void}
     */
    push(val) {
        this.stack.push(val);
        this.cnt.set(val, (this.cnt.get(val) || 0) + 1);
    }

    /**
     * @return {number}
     */
    pop() {
        const maxCnt = Math.max(...this.cnt.values());
        let i = this.stack.length - 1;
        while (this.cnt.get(this.stack[i]) !== maxCnt) {
            i--;
        }
        const val = this.stack.splice(i, 1)[0];
        this.cnt.set(val, this.cnt.get(val) - 1);
        return val;
    }
}

public class FreqStack {
    private Dictionary<int, int> cnt;
    private List<int> stack;

    public FreqStack() {
        cnt = new Dictionary<int, int>();
        stack = new List<int>();
    }

    public void Push(int val) {
        stack.Add(val);
        if (!cnt.ContainsKey(val)) {
            cnt[val] = 0;
        }
        cnt[val]++;
    }

    public int Pop() {
        int maxCnt = 0;
        foreach (var kvp in cnt) {
            maxCnt = Math.Max(maxCnt, kvp.Value);
        }

        for (int i = stack.Count - 1; i >= 0; i--) {
            int val = stack[i];
            if (cnt[val] == maxCnt) {
                cnt[val]--;
                stack.RemoveAt(i);
                return val;
            }
        }

        return -1;
    }
}

type FreqStack struct {
    cnt   map[int]int
    stack []int
}

func Constructor() FreqStack {
    return FreqStack{
        cnt:   make(map[int]int),
        stack: []int{},
    }
}

func (this *FreqStack) Push(val int) {
    this.stack = append(this.stack, val)
    this.cnt[val]++
}

func (this *FreqStack) Pop() int {
    maxCnt := 0
    for _, freq := range this.cnt {
        if freq > maxCnt {
            maxCnt = freq
        }
    }
    for i := len(this.stack) - 1; i >= 0; i-- {
        val := this.stack[i]
        if this.cnt[val] == maxCnt {
            this.cnt[val]--
            this.stack = append(this.stack[:i], this.stack[i+1:]...)
            return val
        }
    }
    return -1
}

class FreqStack() {
    private val cnt = mutableMapOf<Int, Int>()
    private val stack = mutableListOf<Int>()

    fun push(`val`: Int) {
        stack.add(`val`)
        cnt[`val`] = cnt.getOrDefault(`val`, 0) + 1
    }

    fun pop(): Int {
        val maxCnt = cnt.values.maxOrNull() ?: 0
        for (i in stack.lastIndex downTo 0) {
            val v = stack[i]
            if (cnt[v] == maxCnt) {
                cnt[v] = cnt[v]!! - 1
                stack.removeAt(i)
                return v
            }
        }
        return -1
    }
}

class FreqStack {
    private var cnt: [Int: Int]
    private var stack: [Int]

    init() {
        cnt = [:]
        stack = []
    }

    func push(_ val: Int) {
        stack.append(val)
        cnt[val, default: 0] += 1
    }

    func pop() -> Int {
        let maxCnt = cnt.values.max() ?? 0
        for i in stride(from: stack.count - 1, through: 0, by: -1) {
            let val = stack[i]
            if cnt[val] == maxCnt {
                cnt[val]! -= 1
                stack.remove(at: i)
                return val
            }
        }
        return -1
    }
}

Time & Space Complexity

Time complexity:
- $O(1)$ time for each $push()$ function call.
- $O(n)$ time for each $pop()$ function call.
Space complexity: $O(n)$

Where $n$ is the number of elements in the stack.

2. Heap

Intuition

We can use a max-heap to efficiently retrieve the element that should be popped next. Each heap entry stores three pieces of information: the element's frequency, its insertion order (index), and the value itself. By prioritizing higher frequencies and then more recent insertions, the heap always gives us the correct element to pop in logarithmic time.

Algorithm

Initialize a max-heap, a frequency hash map cnt, and an index counter starting at 0.
For push(val): increment the value's frequency, then push a tuple of (frequency, index, val) onto the heap. Increment the index counter.
For pop(): pop the top element from the heap (which has the highest frequency and most recent index among ties), decrement that value's frequency, and return the value.

class FreqStack:

    def __init__(self):
        self.heap = []
        self.cnt = defaultdict(int)
        self.index = 0

    def push(self, val: int) -> None:
        self.cnt[val] += 1
        heapq.heappush(self.heap, (-self.cnt[val], -self.index, val))
        self.index += 1

    def pop(self) -> int:
        _, _, val = heapq.heappop(self.heap)
        self.cnt[val] -= 1
        return val

public class FreqStack {
    private PriorityQueue<int[]> heap;
    private Map<Integer, Integer> cnt;
    private int index;

    public FreqStack() {
        heap = new PriorityQueue<>((a, b) ->
            a[0] != b[0] ? Integer.compare(b[0], a[0]) : Integer.compare(b[1], a[1])
        );
        cnt = new HashMap<>();
        index = 0;
    }

    public void push(int val) {
        cnt.put(val, cnt.getOrDefault(val, 0) + 1);
        heap.offer(new int[]{cnt.get(val), index++, val});
    }

    public int pop() {
        int[] top = heap.poll();
        int val = top[2];
        cnt.put(val, cnt.get(val) - 1);
        return val;
    }
}

class FreqStack {
private:
    priority_queue<vector<int>> heap; // {frequency, index, value}
    unordered_map<int, int> cnt;
    int index;

public:
    FreqStack() : index(0) {}

    void push(int val) {
        cnt[val]++;
        heap.push({cnt[val], index++, val});
    }

    int pop() {
        auto top = heap.top();
        heap.pop();
        int val = top[2];
        cnt[val]--;
        return val;
    }
};

class FreqStack {
    constructor() {
        this.heap = new MaxPriorityQueue({
            priority: (element) => element[0] * 100000 + element[1],
        });
        this.cnt = new Map();
        this.index = 0;
    }

    /**
     * @param {number} val
     * @return {void}
     */
    push(val) {
        this.cnt.set(val, (this.cnt.get(val) || 0) + 1);
        this.heap.enqueue([this.cnt.get(val), this.index++, val]);
    }

    /**
     * @return {number}
     */
    pop() {
        const [, , val] = this.heap.dequeue().element;
        this.cnt.set(val, this.cnt.get(val) - 1);
        return val;
    }
}

public class FreqStack {
    private class Entry : IComparable<Entry> {
        public int Freq { get; }
        public int Index { get; }
        public int Value { get; }

        public Entry(int freq, int index, int value) {
            Freq = freq;
            Index = index;
            Value = value;
        }

        public int CompareTo(Entry other) {
            if (Freq != other.Freq)
                return other.Freq.CompareTo(Freq);
            return other.Index.CompareTo(Index);
        }
    }

    private Dictionary<int, int> cnt;
    private PriorityQueue<Entry, Entry> heap;
    private int index;

    public FreqStack() {
        cnt = new Dictionary<int, int>();
        heap = new PriorityQueue<Entry, Entry>();
        index = 0;
    }

    public void Push(int val) {
        if (!cnt.ContainsKey(val)) {
            cnt[val] = 0;
        }
        cnt[val]++;
        var entry = new Entry(cnt[val], index++, val);
        heap.Enqueue(entry, entry);
    }

    public int Pop() {
        var entry = heap.Dequeue();
        cnt[entry.Value]--;
        return entry.Value;
    }
}

import "container/heap"

type Entry struct {
    freq  int
    index int
    val   int
}

type MaxHeap []Entry

func (h MaxHeap) Len() int { return len(h) }
func (h MaxHeap) Less(i, j int) bool {
    if h[i].freq != h[j].freq {
        return h[i].freq > h[j].freq
    }
    return h[i].index > h[j].index
}
func (h MaxHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *MaxHeap) Push(x interface{}) { *h = append(*h, x.(Entry)) }
func (h *MaxHeap) Pop() interface{} {
    old := *h
    n := len(old)
    x := old[n-1]
    *h = old[:n-1]
    return x
}

type FreqStack struct {
    cnt   map[int]int
    h     *MaxHeap
    index int
}

func Constructor() FreqStack {
    h := &MaxHeap{}
    heap.Init(h)
    return FreqStack{cnt: make(map[int]int), h: h, index: 0}
}

func (this *FreqStack) Push(val int) {
    this.cnt[val]++
    heap.Push(this.h, Entry{this.cnt[val], this.index, val})
    this.index++
}

func (this *FreqStack) Pop() int {
    entry := heap.Pop(this.h).(Entry)
    this.cnt[entry.val]--
    return entry.val
}

import java.util.PriorityQueue

class FreqStack() {
    private val cnt = mutableMapOf<Int, Int>()
    private val heap = PriorityQueue<IntArray> { a, b ->
        if (a[0] != b[0]) b[0] - a[0] else b[1] - a[1]
    }
    private var index = 0

    fun push(`val`: Int) {
        cnt[`val`] = cnt.getOrDefault(`val`, 0) + 1
        heap.offer(intArrayOf(cnt[`val`]!!, index++, `val`))
    }

    fun pop(): Int {
        val top = heap.poll()
        val v = top[2]
        cnt[v] = cnt[v]!! - 1
        return v
    }
}

class FreqStack {
    private var cnt: [Int: Int]
    private var heap: [(freq: Int, index: Int, val: Int)]
    private var index: Int

    init() {
        cnt = [:]
        heap = []
        index = 0
    }

    func push(_ val: Int) {
        cnt[val, default: 0] += 1
        heap.append((cnt[val]!, index, val))
        index += 1
        heap.sort { ($0.freq, $0.index) > ($1.freq, $1.index) }
    }

    func pop() -> Int {
        let top = heap.removeFirst()
        cnt[top.val]! -= 1
        return top.val
    }
}

Time & Space Complexity

Time complexity:
- $O(\log n)$ time for each $push()$ function call.
- $O(\log n)$ time for each $pop()$ function call.
Space complexity: $O(n)$

Where $n$ is the number of elements in the stack.

3. Stack Of Stacks (Hash Map)

Intuition

The key insight is that we can group elements by their frequency level. When an element is pushed for the first time, it goes into stack 1. When pushed again, it also goes into stack 2 (while remaining in stack 1). This way, the stack at the highest frequency level always contains the elements we should consider popping first, and the top of that stack is the most recently pushed among them.

Algorithm

Initialize a frequency hash map cnt, a hash map stacks where each key is a frequency and each value is a stack, and a variable maxCnt to track the current maximum frequency.
For push(val): increment the value's count. If this count exceeds maxCnt, update maxCnt and create a new stack for that frequency. Push the value onto the stack at its new frequency level.
For pop(): pop from the stack at maxCnt, decrement that value's frequency. If the stack at maxCnt becomes empty, decrement maxCnt. Return the popped value.

class FreqStack:

    def __init__(self):
        self.cnt = {}
        self.maxCnt = 0
        self.stacks = {}

    def push(self, val: int) -> None:
        valCnt = 1 + self.cnt.get(val, 0)
        self.cnt[val] = valCnt
        if valCnt > self.maxCnt:
            self.maxCnt = valCnt
            self.stacks[valCnt] = []
        self.stacks[valCnt].append(val)

    def pop(self) -> int:
        res = self.stacks[self.maxCnt].pop()
        self.cnt[res] -= 1
        if not self.stacks[self.maxCnt]:
            self.maxCnt -= 1
        return res

class FreqStack {
    private Map<Integer, Integer> cnt;
    private Map<Integer, Stack<Integer>> stacks;
    private int maxCnt;

    public FreqStack() {
        cnt = new HashMap<>();
        stacks = new HashMap<>();
        maxCnt = 0;
    }

    public void push(int val) {
        int valCnt = cnt.getOrDefault(val, 0) + 1;
        cnt.put(val, valCnt);
        if (valCnt > maxCnt) {
            maxCnt = valCnt;
            stacks.putIfAbsent(valCnt, new Stack<>());
        }
        stacks.get(valCnt).push(val);
    }

    public int pop() {
        int res = stacks.get(maxCnt).pop();
        cnt.put(res, cnt.get(res) - 1);
        if (stacks.get(maxCnt).isEmpty()) {
            maxCnt--;
        }
        return res;
    }
}

class FreqStack {
public:
    unordered_map<int, int> cnt;
    unordered_map<int, stack<int>> stacks;
    int maxCnt;

    FreqStack() {
        maxCnt = 0;
    }

    void push(int val) {
        int valCnt = ++cnt[val];
        if (valCnt > maxCnt) {
            maxCnt = valCnt;
            stacks[valCnt] = stack<int>();
        }
        stacks[valCnt].push(val);
    }

    int pop() {
        int res = stacks[maxCnt].top();
        stacks[maxCnt].pop();
        cnt[res]--;
        if (stacks[maxCnt].empty()) {
            maxCnt--;
        }
        return res;
    }
};

class FreqStack {
    constructor() {
        this.cnt = new Map();
        this.stacks = new Map();
        this.maxCnt = 0;
    }

    /**
     * @param {number} val
     * @return {void}
     */
    push(val) {
        const valCnt = (this.cnt.get(val) || 0) + 1;
        this.cnt.set(val, valCnt);
        if (valCnt > this.maxCnt) {
            this.maxCnt = valCnt;
            if (!this.stacks.has(valCnt)) {
                this.stacks.set(valCnt, []);
            }
        }
        this.stacks.get(valCnt).push(val);
    }

    /**
     * @return {number}
     */
    pop() {
        const res = this.stacks.get(this.maxCnt).pop();
        this.cnt.set(res, this.cnt.get(res) - 1);
        if (this.stacks.get(this.maxCnt).length === 0) {
            this.maxCnt--;
        }
        return res;
    }
}

public class FreqStack {
    private Dictionary<int, int> cnt;
    private Dictionary<int, Stack<int>> stacks;
    private int maxCnt;

    public FreqStack() {
        cnt = new Dictionary<int, int>();
        stacks = new Dictionary<int, Stack<int>>();
        maxCnt = 0;
    }

    public void Push(int val) {
        int valCnt = cnt.ContainsKey(val) ? cnt[val] + 1 : 1;
        cnt[val] = valCnt;

        if (!stacks.ContainsKey(valCnt)) {
            stacks[valCnt] = new Stack<int>();
        }

        stacks[valCnt].Push(val);

        if (valCnt > maxCnt) {
            maxCnt = valCnt;
        }
    }

    public int Pop() {
        int res = stacks[maxCnt].Pop();
        cnt[res]--;

        if (stacks[maxCnt].Count == 0) {
            maxCnt--;
        }

        return res;
    }
}

type FreqStack struct {
    cnt    map[int]int
    stacks map[int][]int
    maxCnt int
}

func Constructor() FreqStack {
    return FreqStack{
        cnt:    make(map[int]int),
        stacks: make(map[int][]int),
        maxCnt: 0,
    }
}

func (this *FreqStack) Push(val int) {
    this.cnt[val]++
    valCnt := this.cnt[val]
    if valCnt > this.maxCnt {
        this.maxCnt = valCnt
        this.stacks[valCnt] = []int{}
    }
    this.stacks[valCnt] = append(this.stacks[valCnt], val)
}

func (this *FreqStack) Pop() int {
    stack := this.stacks[this.maxCnt]
    res := stack[len(stack)-1]
    this.stacks[this.maxCnt] = stack[:len(stack)-1]
    this.cnt[res]--
    if len(this.stacks[this.maxCnt]) == 0 {
        this.maxCnt--
    }
    return res
}

class FreqStack() {
    private val cnt = mutableMapOf<Int, Int>()
    private val stacks = mutableMapOf<Int, ArrayDeque<Int>>()
    private var maxCnt = 0

    fun push(`val`: Int) {
        val valCnt = cnt.getOrDefault(`val`, 0) + 1
        cnt[`val`] = valCnt
        if (valCnt > maxCnt) {
            maxCnt = valCnt
            stacks[valCnt] = ArrayDeque()
        }
        stacks[valCnt]!!.addLast(`val`)
    }

    fun pop(): Int {
        val res = stacks[maxCnt]!!.removeLast()
        cnt[res] = cnt[res]!! - 1
        if (stacks[maxCnt]!!.isEmpty()) {
            maxCnt--
        }
        return res
    }
}

class FreqStack {
    private var cnt: [Int: Int]
    private var stacks: [Int: [Int]]
    private var maxCnt: Int

    init() {
        cnt = [:]
        stacks = [:]
        maxCnt = 0
    }

    func push(_ val: Int) {
        let valCnt = (cnt[val] ?? 0) + 1
        cnt[val] = valCnt
        if valCnt > maxCnt {
            maxCnt = valCnt
            stacks[valCnt] = []
        }
        stacks[valCnt]!.append(val)
    }

    func pop() -> Int {
        let res = stacks[maxCnt]!.removeLast()
        cnt[res]! -= 1
        if stacks[maxCnt]!.isEmpty {
            maxCnt -= 1
        }
        return res
    }
}

Time & Space Complexity

Time complexity:
- $O(1)$ time for each $push()$ function call.
- $O(1)$ time for each $pop()$ function call.
Space complexity: $O(n)$

Where $n$ is the number of elements in the stack.

4. Stack Of Stacks (Dynamic Array)

Intuition

Instead of using a hash map for frequency-indexed stacks, we can use a dynamic array (list). The index in the array represents the frequency level. When an element reaches a new frequency, we extend the array if needed. The last non-empty stack in the array always corresponds to the maximum frequency, eliminating the need to track maxCnt separately.

Algorithm

Initialize a frequency hash map cnt and a list stacks with an empty placeholder at index 0 (since frequencies start at 1).
For push(val): increment the value's count. If this count equals the current length of stacks, append a new empty stack. Push the value onto the stack at index equal to its frequency.
For pop(): pop from the last stack in the list, decrement that value's frequency. If the last stack becomes empty, remove it from the list. Return the popped value.

class FreqStack:

    def __init__(self):
        self.cnt = {}
        self.stacks = [[]]

    def push(self, val: int) -> None:
        valCnt = 1 + self.cnt.get(val, 0)
        self.cnt[val] = valCnt
        if valCnt == len(self.stacks):
            self.stacks.append([])
        self.stacks[valCnt].append(val)

    def pop(self) -> int:
        res = self.stacks[-1].pop()
        self.cnt[res] -= 1
        if not self.stacks[-1]:
            self.stacks.pop()
        return res

public class FreqStack {
    private Map<Integer, Integer> cnt;
    private List<Stack<Integer>> stacks;

    public FreqStack() {
        cnt = new HashMap<>();
        stacks = new ArrayList<>();
        stacks.add(new Stack<>());
    }

    public void push(int val) {
        int valCnt = cnt.getOrDefault(val, 0) + 1;
        cnt.put(val, valCnt);
        if (valCnt == stacks.size()) {
            stacks.add(new Stack<>());
        }
        stacks.get(valCnt).push(val);
    }

    public int pop() {
        Stack<Integer> topStack = stacks.get(stacks.size() - 1);
        int res = topStack.pop();
        cnt.put(res, cnt.get(res) - 1);
        if (topStack.isEmpty()) {
            stacks.remove(stacks.size() - 1);
        }
        return res;
    }
}

class FreqStack {
public:
    unordered_map<int, int> cnt;
    vector<stack<int>> stacks;

    FreqStack() {
        stacks.push_back(stack<int>());
    }

    void push(int val) {
        int valCnt = ++cnt[val];
        if (valCnt == stacks.size()) {
            stacks.push_back(stack<int>());
        }
        stacks[valCnt].push(val);
    }

    int pop() {
        stack<int>& topStack = stacks.back();
        int res = topStack.top();
        topStack.pop();
        if (topStack.empty()) {
            stacks.pop_back();
        }
        cnt[res]--;
        return res;
    }
};

class FreqStack {
    constructor() {
        this.cnt = new Map();
        this.stacks = [[]];
    }

    /**
     * @param {number} val
     * @return {void}
     */
    push(val) {
        const valCnt = (this.cnt.get(val) || 0) + 1;
        this.cnt.set(val, valCnt);
        if (valCnt === this.stacks.length) {
            this.stacks.push([]);
        }
        this.stacks[valCnt].push(val);
    }

    /**
     * @return {number}
     */
    pop() {
        const topStack = this.stacks[this.stacks.length - 1];
        const res = topStack.pop();
        this.cnt.set(res, this.cnt.get(res) - 1);
        if (topStack.length === 0) {
            this.stacks.pop();
        }
        return res;
    }
}

public class FreqStack {
    private Dictionary<int, int> cnt;
    private List<List<int>> stacks;

    public FreqStack() {
        cnt = new Dictionary<int, int>();
        stacks = new List<List<int>>();
        stacks.Add(new List<int>());
    }

    public void Push(int val) {
        int valCnt = cnt.ContainsKey(val) ? cnt[val] + 1 : 1;
        cnt[val] = valCnt;

        if (valCnt == stacks.Count) {
            stacks.Add(new List<int>());
        }

        stacks[valCnt].Add(val);
    }

    public int Pop() {
        int lastIndex = stacks.Count - 1;
        List<int> lastStack = stacks[lastIndex];
        int res = lastStack[lastStack.Count - 1];
        lastStack.RemoveAt(lastStack.Count - 1);

        cnt[res]--;

        if (lastStack.Count == 0) {
            stacks.RemoveAt(lastIndex);
        }

        return res;
    }
}

type FreqStack struct {
    cnt    map[int]int
    stacks [][]int
}

func Constructor() FreqStack {
    return FreqStack{
        cnt:    make(map[int]int),
        stacks: [][]int{{}},
    }
}

func (this *FreqStack) Push(val int) {
    this.cnt[val]++
    valCnt := this.cnt[val]
    if valCnt == len(this.stacks) {
        this.stacks = append(this.stacks, []int{})
    }
    this.stacks[valCnt] = append(this.stacks[valCnt], val)
}

func (this *FreqStack) Pop() int {
    lastStack := this.stacks[len(this.stacks)-1]
    res := lastStack[len(lastStack)-1]
    this.stacks[len(this.stacks)-1] = lastStack[:len(lastStack)-1]
    this.cnt[res]--
    if len(this.stacks[len(this.stacks)-1]) == 0 {
        this.stacks = this.stacks[:len(this.stacks)-1]
    }
    return res
}

class FreqStack() {
    private val cnt = mutableMapOf<Int, Int>()
    private val stacks = mutableListOf(mutableListOf<Int>())

    fun push(`val`: Int) {
        val valCnt = cnt.getOrDefault(`val`, 0) + 1
        cnt[`val`] = valCnt
        if (valCnt == stacks.size) {
            stacks.add(mutableListOf())
        }
        stacks[valCnt].add(`val`)
    }

    fun pop(): Int {
        val lastStack = stacks.last()
        val res = lastStack.removeLast()
        cnt[res] = cnt[res]!! - 1
        if (lastStack.isEmpty()) {
            stacks.removeLast()
        }
        return res
    }
}

class FreqStack {
    private var cnt: [Int: Int]
    private var stacks: [[Int]]

    init() {
        cnt = [:]
        stacks = [[]]
    }

    func push(_ val: Int) {
        let valCnt = (cnt[val] ?? 0) + 1
        cnt[val] = valCnt
        if valCnt == stacks.count {
            stacks.append([])
        }
        stacks[valCnt].append(val)
    }

    func pop() -> Int {
        let res = stacks[stacks.count - 1].removeLast()
        cnt[res]! -= 1
        if stacks[stacks.count - 1].isEmpty {
            stacks.removeLast()
        }
        return res
    }
}

Time & Space Complexity

Time complexity:
- $O(1)$ time for each $push()$ function call.
- $O(1)$ time for each $pop()$ function call.
Space complexity: $O(n)$

Where $n$ is the number of elements in the stack.

Common Pitfalls

Breaking Ties Incorrectly

When multiple elements share the highest frequency, the most recently pushed one must be popped first. A common mistake is to pop an arbitrary element with the maximum frequency, or to pop the first occurrence instead of the last. The heap solution handles this by storing insertion order as a secondary sort key; the stack-of-stacks solution naturally maintains recency by using stack ordering within each frequency level.

Not Decrementing Frequency After Pop

After popping an element, its frequency count must be decremented. Forgetting this step causes the data structure to believe the element still occurs at its previous frequency, leading to incorrect behavior on subsequent operations. Both the frequency map and any frequency-indexed structures must be updated consistently.

Mismanaging the Maximum Frequency Tracker

The stack-of-stacks approach tracks maxCnt to know which stack to pop from. A subtle bug occurs when the stack at maxCnt becomes empty after a pop but maxCnt is not decremented. The next pop would then try to access an empty stack. Always check if the current frequency stack is empty after popping and reduce maxCnt accordingly.

Using a Single Stack and Scanning

While conceptually simple, maintaining a single stack and scanning backwards to find the most frequent recent element leads to O(n) pop operations. This approach times out on large inputs. The key insight is that elements can exist at multiple frequency levels simultaneously (when pushed multiple times), which the stack-of-stacks approach leverages for O(1) operations.

Heap Stale Entries

In the heap-based solution, old entries remain in the heap even after frequencies change through pops. While this doesn't affect correctness (since we decrement the frequency counter separately), it can cause memory bloat. Some implementations add lazy deletion, but the simpler approach is to accept that the heap may contain stale entries whose values no longer reflect current frequencies, relying on the frequency map as the source of truth.