**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 :

` ````
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
```

## View More Similar Problems

## Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →## Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →## Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →## Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →## Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →