# 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]```
```

## Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

## QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

## Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

## Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

## Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w