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:

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#includeusing namespace std; // Function to check if all array// elements is present in the stackbool checkArrInStack(stacks, int arr[],                                  int n){    mapfreq;         // 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 codeint main(){    stacks;    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;         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 approachimport java.util.*;class GFG{ // Function to check if all array// elements is present in the stackstatic boolean checkArrInStack(Stacks,                               int arr[], int n){  HashMapfreq = new HashMap();     // 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 codepublic static void main(String[] args){  Stack s = new Stack();  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 stackdef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{ // Function to check if all array// elements is present in the stackstatic bool checkArrInStack(Stacks,                            int []arr, int n){  Dictionaryfreq = new Dictionary();     // 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 codepublic static void Main(String[] args){  Stack s = new Stack();  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


Output:
YES

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

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