Frequency Stack - Amazon Top Interview Questions


Problem Statement :


Implement a frequency stack with the following methods:

FrequencyStack() constructs a new instance of a frequency stack
append(int val) appends val to the stack
pop() pops and returns the most frequent element in the stack. If there's more than one most frequent element, the one that's closer to the top of the stack should be popped first.
You can assume that for pop, the stack is non-empty when they are called.

Constraints

n ≤ 100,000 where n is the number of methods that will be called to append and pop

Example 1

Input

methods = ["constructor", "append", "append", "append", "pop", "pop", "pop"]

arguments = [[], [1], [1], [2], [], [], []]`

Output

[None, None, None, None, 1, 2, 1]


Solution :



title-img 




                        Solution in C++ :

class FrequencyStack {
    public:
    vector<stack<int>> frequencyToElement;
    unordered_map<int, int> freq;
    FrequencyStack() {
        frequencyToElement.resize(1);
    }

    void append(int val) {
        freq[val]++;
        if (freq[val] >= frequencyToElement.size()) frequencyToElement.push_back({});
        frequencyToElement[freq[val]].push(val);
    }

    int pop() {
        int frequency = frequencyToElement.size() - 1;
        int val = frequencyToElement[frequency].top();
        frequencyToElement[frequency].pop();
        if (frequencyToElement[frequency].size() == 0) frequencyToElement.pop_back();
        freq[val]--;
        return val;
    }
};
                    


                        Solution in Java :

import java.util.*;

class FrequencyStack {
    int i;
    PriorityQueue<Item> pq;
    HashMap<Integer, Integer> sz;
    public FrequencyStack() {
        pq = new PriorityQueue<Item>();
        sz = new HashMap<Integer, Integer>();
        i = 0;
    }

    public void append(int val) {
        int s = 1 + sz.getOrDefault(val, 0);
        pq.add(new Item(val, s, i));
        i++;
        sz.put(val, s);
    }

    public int pop() {
        while (!pq.isEmpty()) {
            Item n = pq.poll();
            int realS = sz.get(n.val);
            if (n.s == realS) {
                sz.put(n.val, realS - 1);
                return n.val;
            }
        }
        return -1;
    }

    static class Item implements Comparable<Item> {
        int val;
        int s;
        long comp;
        public Item(int val, int s, int ind) {
            this.val = val;
            this.s = s;
            comp = 100000L * s + ind;
        }

        public int compareTo(Item i) {
            if (comp > i.comp)
                return -1;
            else if (comp < i.comp)
                return 1;
            else
                return 0;
        }
    }
}
                    


                        Solution in Python : 
                            
class FrequencyStack:
    def __init__(self):
        self.frequencies = defaultdict(int)
        self.heap = []
        self.index = 0

    def append(self, val: int) -> int:
        self.frequencies[val] += 1
        self.index += 1
        heappush(self.heap, (-self.frequencies[val], -self.index, val))

    def pop(self) -> int:
        val = heappop(self.heap)[2]
        self.frequencies[val] -= 1
        return val


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