Problem Description

In this problem, you are given a list of windows, each identified by a number from 1 to n. These windows are stacked on top of each other according to their initial order in the windows array, with the first element at the top and the last at the bottom. You are also provided with a list of queries. Each query requests that a specific window, identified by queries[i], be brought to the top of the stack. The task is to simulate this operation for each query and return the final order of the windows.

Intuition

The goal is to simulate moving specified windows to the top according to the given queries. To achieve the desired final window order efficiently, we can utilize an approach that processes the queries array in reverse. This ensures that windows specified later in the queries will be placed above earlier ones because they will appear earlier as we reverse process the list.

1. Reverse Process Queries: By processing the queries array from the end to the start, we ensure that later queries, which should take precedence, are handled first. For each query, we check if the window has not already been processed and moved to the top; if not, we move it there, recording its existence in a hash table to prevent duplicates.

2. Maintain Final Order: After handling all queries, some windows may not have been queried at all. These windows should remain in their initial relative order but appear below any windows that were moved by queries. Therefore, we append any remaining windows from the original windows list that were not moved during query processing to our result.

This approach, which uses a hash table to track which windows have already been repositioned, ensures that the solution is both efficient and straightforward, maintaining an overall time complexity that depends linearly on the sum of the lengths of windows and queries.

Solution Approach

The solution approach leverages a combination of a hash table and reverse traversal to efficiently reorder the windows based on the queries given. Here's how the algorithm is implemented:

1. Initialize Data Structures:

   • We use a set s to keep track of windows that have already been moved to the top. This helps in avoiding duplicate operations for the same window.
   • An empty list ans is used to store the final order of windows after processing all queries.

2. Reverse Traverse the Queries:

   • We traverse the queries list in reverse. By doing so, we can ensure that any window referred to by a later query will appear above those from earlier queries.
   • For each query q, we check if the window q has not been moved already (i.e., it's not in the set s).
   • If q is not in s, we append it to the ans list and add q to the set s.

3. Append Remaining Windows:

   • After processing all queries, there may be some windows that were never queried. These should remain in the order they exist in the original windows list, but they will be added after those affected by the queries.
   • We iterate over the windows list and append each window to ans if it is not present in the set s.

The final list ans will now contain the windows in the correct order after all query operations:

class Solution:
    def simulationResult(self, windows: List[int], queries: List[int]) -> List[int]:
        s = set()
        ans = []
        for q in queries[::-1]:
            if q not in s:
                ans.append(q)
                s.add(q)
        for w in windows:
            if w not in s:
                ans.append(w)
        return ans

This solution efficiently achieves the desired reordering of windows using a combination of reverse list traversal and a hash table to ensure uniqueness in operations, yielding an effective and concise approach to the problem.

Example Walkthrough

Let's walk through an example to illustrate the solution approach for this problem.

Consider the following inputs:

• Initial windows list: [1, 2, 3, 4, 5]
• Queries: [3, 5, 2, 1]

Step-by-step Execution:

1. Reverse Traverse the Queries:

   • Start with an empty list ans and an empty set s to track windows moved to the top.

   • Traverse queries in reverse order: [1, 2, 5, 3].

   • Process query 1: It's not in the set s, so append 1 to ans, making ans = [1]. Add 1 to s, so s = {1}.

   • Process query 2: Not in s, append 2 to ans, making ans = [1, 2]. Add 2 to s, so s = {1, 2}.

   • Process query 5: Not in s, append 5 to ans, making ans = [1, 2, 5]. Add 5 to s, so s = {1, 2, 5}.

   • Process query 3: Not in s, append 3 to ans, making ans = [1, 2, 5, 3]. Add 3 to s, so s = {1, 2, 3, 5}.

2. Append Remaining Windows:

   • Iterate over the original windows list [1, 2, 3, 4, 5].

   • Append windows not in s to ans in their original order.

   • Window 1: Already in s, skip it.

   • Window 2: Already in s, skip it.

   • Window 3: Already in s, skip it.

   • Window 4: Not in s, append 4 to ans, making ans = [1, 2, 5, 3, 4].

   • Window 5: Already in s, skip it.

Final Result:

The final order of windows is [1, 2, 5, 3, 4], where the queried windows are moved to the top according to the order dictated by reversing the queries, followed by any untouched windows maintaining their initial order.

This example demonstrates the solution's ability to process windows efficiently and ensure the correct order after executing all queries.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n + m), where n is the length of the windows list and m is the length of the queries list. This is because each element in queries is processed once as the loop goes in reverse order, and each element in windows is processed in the second loop.

The space complexity is O(m), as the set s used to store unique elements is primarily influenced by the number of elements in queries, which can be at most m.

Learn more about how to find time and space complexity quickly.