# Count the number of pop operations on stack to get each element of the array

Last Updated : 15 Mar, 2023

Prerequisite: Stack, Hashing

Given a stack of N numbers and an array of numbers. Count the number of pop operations required to get each element of the array. Once an element is popped then it’s not pushed back again. Assume that all the elements from the array are present inside the stack initially.

Examples:

Input: N = 5, Stack: 6 4 3 2 1 ,  Array: 6 3 4 1 2
Output: 1 2 0 2 0
The 1st element of the stack is the same as the array elements. So to get 6, one pop is required i.e. pop 6 from the stack. To get 3, 2 elements will be popped i.e. 4 and 3. To get 4, 4 was already popped, thus we won’t pop more elements. Similarly, to get 1, we will pop 2 elements and for the 2, no pop count will be added.

Approach: This question can be solved easily by using a stack. We will keep popping the elements till we find the element we are searching for. The only hurdle is how to handle the case when the element is already popped and is not present in the stack. For that, we will maintain a hash map. As we pop an element from the stack we will insert that element in the hash map so that if the element comes later in the array we will first check if it’s present inside the hash maps or in other words has been popped out from the stack previously. Otherwise, we will know it’s present inside the stack and we will start popping the elements till we find the required number.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement above approach` `#include ` `using` `namespace` `std;`   `// Function to find the count` `void` `countEle(stack<``int``>& s, ``int` `a[], ``int` `N)` `{` `    ``// Hashmap to store all the elements` `    ``// which are popped once.` `    ``unordered_map<``int``, ``bool``> mp;` `    ``for` `(``int` `i = 0; i < N; ++i) {` `        ``int` `num = a[i];`   `        ``// Check if the number is present` `        ``// in the hashmap Or in other words` `        ``// been popped out from the stack before.` `        ``if` `(mp.find(num) != mp.end())` `            ``cout << ``"0 "``;`   `        ``else` `{` `            ``int` `cnt = 0;`   `            ``// Keep popping the elements` `            ``// while top is not equal to num` `            ``while` `(s.top() != num) {` `                ``mp[s.top()] = ``true``;` `                ``s.pop();` `                ``cnt++;` `            ``}` `            ``// Pop the top ie. equal to num` `            ``s.pop();` `            ``cnt++;`   `            ``// Print the number of elements popped.` `            ``cout << cnt << ``" "``;` `        ``}` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``int` `N = 5;`   `    ``stack<``int``> s;` `    ``s.push(1);` `    ``s.push(2);` `    ``s.push(3);` `    ``s.push(4);` `    ``s.push(6);`   `    ``int` `a[] = { 6, 3, 4, 1, 2 };` `    ``countEle(s, a, N);`   `    ``return` `0;` `}`

## Java

 `// Java program to implement above approach ` `import` `java.util.HashMap;` `import` `java.util.Stack;`   `class` `GFG ` `{`   `    ``// Function to find the count` `    ``public` `static` `void` `countEle(Stack s, ` `                                ``int``[] a, ``int` `N) ` `    ``{`   `        ``// Hashmap to store all the elements` `        ``// which are popped once.` `        ``HashMap mp = ``new` `HashMap<>();` `        ``for` `(``int` `i = ``0``; i < N; ++i)` `        ``{` `            ``int` `num = a[i];`   `            ``// Check if the number is present` `            ``// in the hashmap Or in other words` `            ``// been popped out from the stack before.` `            ``if` `(mp.containsKey(num))` `                ``System.out.print(``"0 "``);` `            ``else` `            ``{` `                ``int` `cnt = ``0``;`   `                ``// Keep popping the elements` `                ``// while top is not equal to num` `                ``while` `(s.peek() != num) ` `                ``{` `                    ``mp.put(s.peek(), ``true``);` `                    ``s.pop();` `                    ``cnt++;` `                ``}`   `                ``// Pop the top ie. equal to num` `                ``s.pop();` `                ``cnt++;`   `                ``// Print the number of elements popped.` `                ``System.out.print(cnt + ``" "``);` `            ``}` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``int` `N = ``5``;`   `        ``Stack s = ``new` `Stack<>();` `        ``s.add(``1``);` `        ``s.add(``2``);` `        ``s.add(``3``);` `        ``s.add(``4``);` `        ``s.add(``6``);`   `        ``int``[] a = { ``6``, ``3``, ``4``, ``1``, ``2` `};` `        ``countEle(s, a, N);` `    ``}` `}`   `// This code is contributed by` `// sanjeev2552`

## Python3

 `# Python3 program to implement above approach `   `# Function to find the count ` `def` `countEle(s, a, N): ` ` `  `    ``# Hashmap to store all the elements ` `    ``# which are popped once. ` `    ``mp ``=` `{} ` `    ``for` `i ``in` `range``(``0``, N):  ` `        ``num ``=` `a[i] `   `        ``# Check if the number is present ` `        ``# in the hashmap Or in other words ` `        ``# been popped out from the stack before. ` `        ``if` `num ``in` `mp: ` `            ``print``(``"0"``, end ``=` `" "``) `   `        ``else``:` `            ``cnt ``=` `0`   `            ``# Keep popping the elements ` `            ``# while top is not equal to num ` `            ``while` `s[``-``1``] !``=` `num:` `                ``mp[s.pop()] ``=` `True` `                ``cnt ``+``=` `1` `             `  `            ``# Pop the top ie. equal to num ` `            ``s.pop()` `            ``cnt ``+``=` `1`   `            ``# Print the number of elements popped. ` `            ``print``(cnt, end ``=` `" "``) `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:` ` `  `    ``N ``=` `5` `    ``s ``=` `[] ` `    ``s.append(``1``) ` `    ``s.append(``2``) ` `    ``s.append(``3``) ` `    ``s.append(``4``) ` `    ``s.append(``6``) `   `    ``a ``=` `[``6``, ``3``, ``4``, ``1``, ``2``]  ` `    ``countEle(s, a, N) `   `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the above approach: ` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG ` `{`   `    ``// Function to find the count` `    ``public` `static` `void` `countEle(Stack<``int``> s, ` `                                ``int``[] a, ``int` `N) ` `    ``{`   `        ``// Hashmap to store all the elements` `        ``// which are popped once.` `        ``Dictionary<``int``,` `                   ``bool``> mp = ``new` `Dictionary<``int``,` `                                             ``bool``>();` `        ``for` `(``int` `i = 0; i < N; ++i)` `        ``{` `            ``int` `num = a[i];`   `            ``// Check if the number is present` `            ``// in the hashmap Or in other words` `            ``// been popped out from the stack before.` `            ``if` `(mp.ContainsKey(num))` `                ``Console.Write(``"0 "``);` `            ``else` `            ``{` `                ``int` `cnt = 0;`   `                ``// Keep popping the elements` `                ``// while top is not equal to num` `                ``while` `(s.Peek() != num) ` `                ``{` `                    ``mp.Add(s.Peek(), ``true``);` `                    ``s.Pop();` `                    ``cnt++;` `                ``}`   `                ``// Pop the top ie. equal to num` `                ``s.Pop();` `                ``cnt++;`   `                ``// Print the number of elements popped.` `                ``Console.Write(cnt + ``" "``);` `            ``}` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args) ` `    ``{` `        ``int` `N = 5;`   `        ``Stack<``int``> s = ``new` `Stack<``int``>();` `        ``s.Push(1);` `        ``s.Push(2);` `        ``s.Push(3);` `        ``s.Push(4);` `        ``s.Push(6);`   `        ``int``[] a = { 6, 3, 4, 1, 2 };` `        ``countEle(s, a, N);` `    ``}` `}`   `// This code is contributed by PrinciRaj1992 `

## Javascript

 ``

Output

`1 2 0 2 0 `

Complexity Analysis:

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

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