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Check if all array elements are present in a given stack or not

  • Last Updated : 10 Jun, 2021

Given a stack of integers S and an array of integers arr[], the task is to check if all the array elements are present in the stack or not.

Examples:

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Input: S = {10, 20, 30, 40, 50}, arr[] = {20, 30}
Output: Yes
Explanation:
Elements 20 and 30 are present in the stack.



Input: S = {50, 60}, arr[] = {60, 50}
Output: Yes
Explanation:
Elements 50 and 60 are present in the stack.

Approach: The idea is to maintain the frequency of array elements in a Hashmap. Now, while the stack is not empty, keep popping the elements out from the stack and reduce the frequency of elements from the Hashmap. Finally, when the stack is empty, check that the frequency of every element in the hash-map is zero or not. If found to be true, print Yes. Otherwise, print No.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to check if all array
// elements is present in the stack
bool checkArrInStack(stack<int>s, int arr[],
                                  int n)
{
    map<int, int>freq;
     
    // Store the frequency
    // of array elements
    for(int i = 0; i < n; i++)
        freq[arr[i]]++;
         
    // Loop while the elements in the
    // stack is not empty
    while (!s.empty())
    {
        int poppedEle = s.top();
        s.pop();
         
        // Condition to check if the
        // element is present in the stack
        if (freq[poppedEle])
            freq[poppedEle] -= 1;
    }
    if (freq.size() == 0)
        return 0;
         
    return 1;
}
 
// Driver code
int main()
{
    stack<int>s;
    s.push(10);
    s.push(20);
    s.push(30);
    s.push(40);
    s.push(50);
     
    int arr[] = {20, 30};
     
    int n = sizeof arr / sizeof arr[0];
     
    if (checkArrInStack(s, arr, n))
        cout << "YES\n";
    else
        cout << "NO\n";
}
 
// This code is contributed by Stream_Cipher

Java




// Java program of
// the above approach
import java.util.*;
class GFG{
 
// Function to check if all array
// elements is present in the stack
static boolean checkArrInStack(Stack<Integer>s,
                               int arr[], int n)
{
  HashMap<Integer,
          Integer>freq = new HashMap<Integer,
                                     Integer>();
   
  // Store the frequency
  // of array elements
  for(int i = 0; i < n; i++)
    if(freq.containsKey(arr[i]))
      freq.put(arr[i], freq.get(arr[i]) + 1);
  else
    freq.put(arr[i], 1);
 
  // Loop while the elements in the
  // stack is not empty
  while (!s.isEmpty())
  {
    int poppedEle = s.peek();
    s.pop();
 
    // Condition to check if the
    // element is present in the stack
    if (freq.containsKey(poppedEle))
      freq.put(poppedEle, freq.get(poppedEle) - 1);
  }
  if (freq.size() == 0)
    return false;
  return true;
}
 
// Driver code
public static void main(String[] args)
{
  Stack<Integer> s = new Stack<Integer>();
  s.add(10);
  s.add(20);
  s.add(30);
  s.add(40);
  s.add(50);
 
  int arr[] = {20, 30};
  int n = arr.length;
 
  if (checkArrInStack(s, arr, n))
    System.out.print("YES\n");
  else
    System.out.print("NO\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python program of
# the above approach
 
# Function to check if all array
# elements is present in the stack
def checkArrInStack(s, arr):
  freq = {}
   
  # Store the frequency
  # of array elements
  for ele in arr:
    freq[ele] = freq.get(ele, 0) + 1
   
  # Loop while the elements in the
  # stack is not empty
  while s:
    poppedEle = s.pop()
     
    # Condition to check if the
    # element is present in the stack
    if poppedEle in freq:
      freq[poppedEle] -= 1
       
      if not freq[poppedEle]:
        del freq[poppedEle]
  if not freq:
    return True
  return False
 
# Driver Code
if __name__ == "__main__":
  s = [10, 20, 30, 40, 50]
  arr = [20, 30]
   
  if checkArrInStack(s, arr):
    print("YES")
  else:
    print("NO")
      

C#




// C# program of
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to check if all array
// elements is present in the stack
static bool checkArrInStack(Stack<int>s,
                            int []arr, int n)
{
  Dictionary<int,
             int>freq = new Dictionary<int,
                                       int>();
   
  // Store the frequency
  // of array elements
  for(int i = 0; i < n; i++)
    if(freq.ContainsKey(arr[i]))
      freq[arr[i]] = freq[arr[i]] + 1;
  else
    freq.Add(arr[i], 1);
 
  // Loop while the elements in the
  // stack is not empty
  while (s.Count != 0)
  {
    int poppedEle = s.Peek();
    s.Pop();
 
    // Condition to check if the
    // element is present in the stack
    if (freq.ContainsKey(poppedEle))
      freq[poppedEle] = freq[poppedEle] - 1;
  }
  if (freq.Count == 0)
    return false;
  return true;
}
 
// Driver code
public static void Main(String[] args)
{
  Stack<int> s = new Stack<int>();
  s.Push(10);
  s.Push(20);
  s.Push(30);
  s.Push(40);
  s.Push(50);
  int []arr = {20, 30};
  int n = arr.Length;
 
  if (checkArrInStack(s, arr, n))
    Console.Write("YES\n");
  else
    Console.Write("NO\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program of the above approach
 
// Function to check if all array
// elements is present in the stack
function checkArrInStack(s, arr, n)
{
    var freq = new Map();
     
    // Store the frequency
    // of array elements
    for(var i = 0; i < n; i++)
    {
        if(freq.has(arr[i]))
            freq.set(arr[i], freq.get(arr[i])+1)
        else   
            freq.set(arr[i], 1)
    }
         
    // Loop while the elements in the
    // stack is not empty
    while (s.length!=0)
    {
        var poppedEle = s[s.length-1];
        s.pop();
         
        // Condition to check if the
        // element is present in the stack
        if (freq.has(poppedEle))
            freq.set(poppedEle, freq.get(poppedEle)-1);
    }
    if (freq.size == 0)
        return 0;
         
    return 1;
}
 
// Driver code
var s = [];
s.push(10);
s.push(20);
s.push(30);
s.push(40);
s.push(50);
 
var arr = [20, 30];
 
var n = arr.length;
 
if (checkArrInStack(s, arr, n))
    document.write( "YES");
else
    document.write( "NO");
 
 
 
</script>
Output: 
YES

Time Complexity: O(N)
Auxiliary Space: O(N)




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