Maximum Stack - Microsoft Top Interview Questions


Problem Statement :


Implement a maximum stack with the following methods:

MaximumStack() constructs a new instance of a maximum stack

append(int val) appends val to the stack

peek() retrieves the last element in the stack
max() retrieves the maximum value in the stack

pop() pops and returns the last element in the stack

popmax() pops and returns the last occurrence of the maximum value in the stack

You can assume that for peek, max, pop and popmax, the stack is non-empty when they are called.



Constraints



n ≤ 100,000 where n is the number of methods that will be called.

Example 1

Input

methods = ["constructor", "append", "append", "append", "peek", "max", "popmax", "max", "pop", "peek"]

arguments = [[], [1], [3], [2], [], [], [], [], [], []]`

Output

[None, None, None, None, 2, 3, 3, 2, 2, 1]

Explanation

We create a MaximumStack

Append 1 to the stack

Append 3 to the stack

Append 2 to the stack

Peek the top element which is 2

Max value of the stack so far is max(1, 3, 2) == 3

We pop the max value of 3

Max value of the stack is now max(1, 2) == 2

We pop the top element which is 2

Peek the top element which is now 1



Solution :



                        Solution in C++ :

class MaximumStack {
    public:
    int idx = 0;
    set<pair<int, int>> s;
    vector<pair<int, int>> v;

    void append(int val) {
        v.emplace_back(val, ++idx);
        s.emplace(val, idx);
    }

    int peek() {
        while (s.find(v.back()) == s.end()) v.pop_back();
        return v.back().first;
    }

    int max() {
        return s.rbegin()->first;
    }

    int pop() {
        while (s.find(v.back()) == s.end()) v.pop_back();
        auto [x, y] = v.back();
        v.pop_back();
        s.erase(s.find({x, y}));
        return x;
    }

    int popmax() {
        auto [x, y] = *s.rbegin();
        s.erase(prev(s.end()));
        return x;
    }
};
                    

                        Solution in Java :

class MaximumStack {
    public:
    map<int, vector<list<int>::iterator>> m1;
    list<int> L;
    MaximumStack() {
    }
    void append(int val) {
        L.push_back(val);
        m1[val].push_back(prev(L.end()));
    }

    int peek() {
        return *prev(L.end());
    }
    int max() {
        return m1.rbegin()->first;
    }
    int pop() {
        int val = peek();
        erase(val);
        return val;
    }
    void erase(int val) {
        vector<list<int>::iterator> &v = m1[val];
        auto it = v.back();
        v.pop_back();
        if (v.size() == 0) m1.erase(val);
        L.erase(it);
    }
    int popmax() {
        int val = m1.rbegin()->first;
        erase(val);
        return val;
    }
};
                    

                        Solution in Python : 
                            
class MaximumStack:
    def __init__(self):
        # `time` is increased when an element is appended. So each `time`
        # corresponds to a unique element, as recorded in `time_to_val`.
        # When an element has been popped, its corresponding time, thus
        # becoming invalid, will be deleted from `time_to_val`.
        self.time = 0
        self.time_to_val = {}

        # a LIFO stack of uniq pairs `(element, time)`
        self.stk = []

        # a priority queue of uniq pairs `(-element, -time)` for max-related stuff
        self.pq = []

    def append(self, val):
        self.time += 1
        self.time_to_val[self.time] = val
        self.stk.append((val, self.time))
        heapq.heappush(self.pq, (-val, -self.time))

    # After this operation, the top elements of `stk` and `pq` are valid.
    def _pop_invalid_times(self, invalid_time):
        del self.time_to_val[invalid_time]
        while self.stk and (self.stk[-1][1] not in self.time_to_val):
            self.stk.pop()
        while self.pq and (-self.pq[0][1] not in self.time_to_val):
            heapq.heappop(self.pq)

    def peek(self):
        return self.stk[-1][0]

    def max(self):
        return -self.pq[0][0]

    def pop(self):
        entry = self.stk.pop()
        self._pop_invalid_times(entry[1])
        return entry[0]

    def popmax(self):
        entry = heapq.heappop(self.pq)
        self._pop_invalid_times(-entry[1])
        return -entry[0]
                    

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