Design a stack with max frequency element operation

Last Updated : 02 Nov, 2024

Given n elements and the task is to implement a stack which removes and returns the maximum frequency element on every pop operation. If there's a tie in the frequency then the topmost highest frequency element will be returned.
Examples:

Input:
push(4) 8
push(6) 6
push(7) 7
push(6) 6
push(8); 4
Output:
pop() -> returns 6, as 6 is the most frequent (frequency of 6 = 2 ).
pop() -> returns 8 (6 also has the highest frequency but it is not the topmost)

Naive Approach - Priority Queue and Hash Map

The idea is to use priority queue to order elements by frequency and a hash map to store frequencies of items.

push(x) : We increment frequency in map and insert new frequency in the priority queue.
pop(): We pop the top item from priority queue (which is the max frequency item).

C++

#include <bits/stdc++.h>
using namespace std;

priority_queue<pair<int, pair<int, int> > > q;
unordered_map<int, int> freq;
int pos = 0;

void push(int x)
{
    ++freq[x];
    q.emplace(freq[x], make_pair(pos, x));
    ++pos;
}

int pop()
{
    auto val = q.top();
    q.pop();
    int x = val.second.second;
    freq[x]--;
    return x;
}

int main()
{
    push(4);
    push(6);
    push(7);
    push(6);
    push(8);

    cout << (pop()) << "\n";
    cout << (pop());

    return 0;
}

Java

import java.util.PriorityQueue;
import java.util.HashMap;

public class Main {

    // PriorityQueue to store elements in ascending order of their frequency
    // Each element of PriorityQueue is a triple of (-frequency, -position, value)
    static PriorityQueue<int[]> q = new PriorityQueue<>(
            (a, b) -> {
                if (a[0] != b[0]) {
                    return Integer.compare(a[0], b[0]);
                }
                if (a[1] != b[1]) {
                    return Integer.compare(a[1], b[1]);
                }
                return Integer.compare(a[2], b[2]);
            }
    );

    // HashMap to store the frequency of each element
    static HashMap<Integer, Integer> freq = new HashMap<>();
    static int pos = 0;

    // Function to push an element into PriorityQueue and update its frequency
    static void push(int x) {
      
        // Push a triple of (-frequency, -position, value) into PriorityQueue
        q.offer(new int[]{-freq.getOrDefault(x, 0), -pos, x});
      
        // Increment the frequency of element x in HashMap
        freq.put(x, freq.getOrDefault(x, 0) + 1);
      
        // Increment the position
        pos++;
    }

    // Function to remove an element from PriorityQueue and update its frequency
    static int pop() {
      
        // Remove the top element from PriorityQueue
        int[] val = q.poll();
      
        // Get the value of the removed element
        int x = val[2];
      
        // Decrement the frequency of element x in HashMap
        freq.put(x, freq.get(x) - 1);
      
        return x;
    }

    public static void main(String[] args) {
        push(4);
        push(6);
        push(7);
        push(6);
        push(8);
        System.out.println(pop());
        System.out.println(pop());
    }
}

Python

import heapq
from collections import defaultdict

q = []
freq = defaultdict(int)
pos = 0


def push(x):
    global pos
    heapq.heappush(q, (-freq[x], -pos, x))
    freq[x] += 1
    pos += 1


def pop():
    x = heapq.heappop(q)[2]
    freq[x] -= 1
    return x


push(4)
push(6)
push(7)
push(6)
push(8)

# Pop elements
print(pop())
print(pop())

using System;
using System.Collections.Generic;

class FrequencyStack
{
    private Dictionary<int, Stack<int>> freqStack = new Dictionary<int, Stack<int>>();
    private Dictionary<int, int> freq = new Dictionary<int, int>();
    private int maxFreq = 0;

    public void Push(int x)
    {
        if (!freq.ContainsKey(x))
            freq[x] = 0;
        
        freq[x]++;
        if (!freqStack.ContainsKey(freq[x]))
            freqStack[freq[x]] = new Stack<int>();
        
        freqStack[freq[x]].Push(x);
        if (freq[x] > maxFreq)
            maxFreq = freq[x];
    }

    public int Pop()
    {
        int x = freqStack[maxFreq].Pop();
        if (freqStack[maxFreq].Count == 0)
            maxFreq--;

        freq[x]--;
        if (freq[x] == 0)
            freq.Remove(x);

        return x;
    }
}

class Program
{
    static void Main()
    {
        FrequencyStack stack = new FrequencyStack();
        stack.Push(4);
        stack.Push(6);
        stack.Push(7);
        stack.Push(6);
        stack.Push(8);

        // Pop elements
        Console.WriteLine(stack.Pop());
        Console.WriteLine(stack.Pop());
    }
}

JavaScript

// Create an empty object to store frequencies
let freq = {};

// Create an empty array to store values along with their frequencies and positions
let arr = [];

// Create a variable to keep track of the position of the last added element
let pos = 0;

// Function to add an element to the array and update its frequency
function push(x) {

// If the frequency of the element is not already in the object, add it
if (freq[x] === undefined) freq[x] = 0;

// Increase the frequency of the element and store its frequency, position and value in the array
freq[x]++;
arr.push([freq[x], ++pos, x]);
}

// Function to remove the element with the highest frequency and position from the array
function pop() {

// Create variables to store the maximum frequency and position of elements in the array
let maxIdx = -1;
let maxVal = -1;

// Loop through the array to find the element with the highest frequency and position
for (let i = 0; i < arr.length; i++) {
let [freq, idx, val] = arr[i];
if (freq > maxVal || (freq == maxVal && idx > maxIdx)) {
maxIdx = idx;
maxVal = freq;
}
}

// Loop through the array again to find the index of the element with the highest frequency and position
let idx = -1;
for (let i = 0; i < arr.length; i++) {
if (arr[i][1] == maxIdx) {
idx = i;
break;
}
}

// Store the frequency, position and value of the element to be removed
let [freq, pos, x] = arr[idx];

// Decrease the frequency of the element in the object
freq--;

// If the frequency of the element becomes zero, delete it from the object
if (freq == 0) delete freq[x];
else freq[x] = freq;

// Remove the element from the array
arr.splice(idx, 1);

// Return the value of the removed element
return x;
}

// Add some elements to the array
push(4);
push(6);
push(7);
push(6);
push(8);

// Remove and print the two elements with the highest frequency and position
console.log(pop());
console.log(pop());

Output

6
8

Time Complexity: O(Log (n)) on average'
Auxiliary Space: O(n)

Expected Approach - Using Two Hash Maps or Dictionaries

One is frequency Hash Map which maps elements to their frequencies and other is setMap which maps all the element with same frequency in one group (Stack).
FrequencyStack has 2 functions:

push(int x): map the element (x) with frequency HashMap and update the maxfreq variable ( i.e. holds the maximum frequency till now ). setMap maintains a stack which contains all the elements with same frequency.
pop(): First get the maxfreq element from setMap and then decrement the frequency of the popped element. After popping, if the stack becomes empty then decrement the maxfreq.

C++

// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;

// freqMap is to map element to its frequency
unordered_map<int, int> freqMap;

// setMap is to map frequency to the
// element list with this frequency
unordered_map<int, stack<int> > setMap;

// Variable which stores maximum frequency
// of the stack element
int maxfreq = 0;

// Function to insert x in the stack
void push(int x)
{

    // Frequency of x
    int freq = freqMap[x] + 1;

    // Mapping of x with its frequency
    freqMap[x]= freq;

    // Update maxfreq variable
    if (freq > maxfreq)
        maxfreq = freq;

    // Map element to its frequency list
    // If given frequency list doesn't exist
    // make a new list of the required frequency
    setMap[freq].push(x);
}

// Function to remove maximum frequency element
int pop()
{

    // Remove element from setMap
    // from maximum frequency list
    int top = setMap[maxfreq].top();
    setMap[maxfreq].pop();

    // Decrement the frequency of the popped element
    freqMap[top]--;

    // If whole list is popped
    // then decrement the maxfreq
    if (setMap[maxfreq].size() == 0)
        maxfreq--;
    return top;
}

// Driver code
int main()
{
    push(4);
    push(6);
    push(7);
    push(6);
    push(8);
    cout << (pop()) << "\n" ;
    cout << (pop());
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

public class freqStack {

    // freqMap is to map element to its frequency
    static Map<Integer, Integer> freqMap = new HashMap<>();

    // setMap is to map frequency to the
    // element list with this frequency
    static Map<Integer, Stack<Integer> > setMap = new HashMap<>();

    // Variable which stores maximum frequency
    // of the stack element
    static int maxfreq = 0;

    // Function to insert x in the stack
    public static void push(int x)
    {

        // Frequency of x
        int freq = freqMap.getOrDefault(x, 0) + 1;

        // Mapping of x with its frequency
        freqMap.put(x, freq);

        // Update maxfreq variable
        if (freq > maxfreq)
            maxfreq = freq;

        // Map element to its frequency list
        // If given frequency list doesn't exist
        // make a new list of the required frequency
        setMap.computeIfAbsent(freq, z -> new Stack()).push(x);
    }

    // Function to remove maximum frequency element
    public static int pop()
    {

        // Remove element from setMap
        // from maximum frequency list
        int top = setMap.get(maxfreq).pop();

        // Decrement the frequency of the popped element
        freqMap.put(top, freqMap.get(top) - 1);

        // If whole list is popped
        // then decrement the maxfreq
        if (setMap.get(maxfreq).size() == 0)
            maxfreq--;
        return top;
    }

    // Driver code
    public static void main(String[] args)
    {

        // Push elements to the stack
        push(4);
        push(6);
        push(7);
        push(6);
        push(8);

        // Pop elements
        System.out.println(pop());
        System.out.println(pop());
    }
}

Python

# Python3 implementation of the approach 

# freqMap is to map element to its frequency 
freqMap = {}; 

# setMap is to map frequency to the 
# element list with this frequency 
setMap = {}; 

# Variable which stores maximum frequency 
# of the stack element 
maxfreq = 0; 

# Function to insert x in the stack 
def push(x) :
    global maxfreq;
    if x not in freqMap :
        freqMap[x] = 0
        
    # Frequency of x
    freq = freqMap[x] + 1;
    
    # Mapping of x with its Frequency
    freqMap[x]= freq
    
    # Update maxfreq Variable
    if (freq > maxfreq) :
        maxfreq = freq
        
    # Map element to its frequency list
    # If given frequency list doesn't exist
    # make a new list of the required frequency
    if freq not in setMap :
        setMap[freq] = []
    
    setMap[freq].append(x); 

# Function to remove maximum frequency element 
def pop() :

    global maxfreq
    
    # Remove element from setMap
    # from maximum frequency list
    top = setMap[maxfreq][-1];
    setMap[maxfreq].pop();
    
    # Decrement the frequency
    # of the popped element
    freqMap[top] -= 1;
    
    # If whole list is popped
    # then decrement the maxfreq
    if (len(setMap[maxfreq]) == 0) :
        maxfreq -= 1;
        
    return top; 

# Driver code 
if __name__ == "__main__" : 

    # Push elements to the stack 
    push(4); 
    push(6); 
    push(7); 
    push(6); 
    push(8); 

    # Pop elements 
    print(pop()) ; 
    print(pop()); 

# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
    
    // freqMap is to map element to its frequency
    static Dictionary<int, int> freqMap = new Dictionary<int, int>();
 
    // setMap is to map frequency to the
    // element list with this frequency
    static Dictionary<int, Stack<int>> setMap = new Dictionary<int, Stack<int>>();
 
    // Variable which stores maximum frequency
    // of the stack element
    static int maxfreq = 0;
 
    // Function to insert x in the stack
    static void push(int x)
    {
        int freq = 1;
        
        // Frequency of x
        if(freqMap.ContainsKey(x))
        {
            freq = freq + freqMap[x];
        }
 
        // Mapping of x with its frequency
        freqMap[x] = freq;
 
        // Update maxfreq variable
        if (freq > maxfreq)
            maxfreq = freq;
 
        // Map element to its frequency list
        // If given frequency list doesn't exist
        // make a new list of the required frequency
        if(!setMap.ContainsKey(freq))
        {
            setMap[freq] = new Stack<int>();
        }
     
        setMap[freq].Push(x);
    }
 
    // Function to remove maximum frequency element
    static int pop()
    {
 
        // Remove element from setMap
        // from maximum frequency list
        int top = setMap[maxfreq].Peek();
        setMap[maxfreq].Pop();
 
        // Decrement the frequency of the popped element
        freqMap[top] = freqMap[top] - 1;
 
        // If whole list is popped
        // then decrement the maxfreq
        if (setMap[maxfreq].Count == 0)
            maxfreq--;
        return top;
    }
    
  static void Main() {
    // Push elements to the stack
    push(4);
    push(6);
    push(7);
    push(6);
    push(8);
 
    // Pop elements
    Console.WriteLine(pop()) ;
    Console.WriteLine(pop());
  }
}

// This code is contributed by rameshtravel07.

JavaScript

<script>

// Javascript implementation of the approach

// freqMap is to map element to its frequency
var freqMap = new Map();

// setMap is to map frequency to the
// element list with this frequency
var setMap = new Map();

// Variable which stores maximum frequency
// of the stack element
var maxfreq = 0;

// Function to insert x in the stack
function push(x)
{

    // Frequency of x
    if(!freqMap.has(x))
        freqMap.set(x, 1)
    else
        freqMap.set(x, freqMap.get(x)+1)
    
    var freq = freqMap.get(x)

    // Mapping of x with its frequency
    freqMap.set(x, freq);

    // Update maxfreq variable
    if (freq > maxfreq)
        maxfreq = freq;

    // Map element to its frequency list
    // If given frequency list doesn't exist
    // make a new list of the required frequency
    if(!setMap.has(freq))
    {
        setMap.set(freq, [x])
    }
    else
    {
        var tmp = setMap.get(freq);
        tmp.push(x);
        setMap.set(freq, tmp);
    }
}

// Function to remove maximum frequency element
function pop()
{

    // Remove element from setMap
    // from maximum frequency list
    var tmp = setMap.get(maxfreq);
    var top = tmp[tmp.length-1];
    tmp.pop();
    setMap.set(maxfreq, tmp);

    // Decrement the frequency of the popped element
    if(freqMap.has(top))
        freqMap.set(top, freqMap.get(top)-1)

    // If whole list is popped
    // then decrement the maxfreq
    if (setMap.get(maxfreq).length == 0)
        maxfreq--;
    return top;
}

// Driver code
// Push elements to the stack
push(4);
push(6);
push(7);
push(6);
push(8);

// Pop elements
document.write( (pop()) + "<br>");
document.write( (pop()));

// This code is contributed by itsok.
</script>

Output

6
8

Time Complexity: O(1) on average for both push() and pop()
Auxiliary Space: O(n)

Design an efficient data structure for given operations

dekay

Improve

Article Tags :

Practice Tags :

Design a stack with max frequency element operation

Naive Approach - Priority Queue and Hash Map

Expected Approach - Using Two Hash Maps or Dictionaries

Similar Reads

Thank You!

What kind of Experience do you want to share?