Problem Description

You have n windows open, numbered from 1 to n. These windows are arranged in a stack-like order where you can navigate between them using alt + tab functionality.

You're given two arrays:

• windows: Contains the initial order of windows from top to bottom (first element is at the top, last element is at the bottom)
• queries: Contains a sequence of window numbers that will be brought to the top one by one

When a window in queries[i] is brought to the top, it moves from its current position to become the topmost window, while all other windows maintain their relative order.

Your task is to simulate this process and return the final arrangement of windows after all queries have been processed.

For example, if you have windows [1, 2, 3] and queries [3, 2]:

• After query 3: Window 3 moves to top → [3, 1, 2]
• After query 2: Window 2 moves to top → [2, 3, 1]

The solution cleverly processes queries in reverse order. Since the last query determines which window ends up on top, by traversing backwards and using a hash table to track already-placed windows, we can build the final arrangement efficiently. Windows from queries that appear later are placed first (as they should be on top), followed by any remaining windows from the original arrangement that weren't queried.

Intuition

The key insight comes from recognizing that the final position of a window depends on when it was last queried. If we process queries in the normal forward order, we'd need to constantly rearrange windows, which is inefficient.

Think about it this way: if a window appears multiple times in queries, only its last appearance matters for its final position. For instance, if window 2 is queried at positions i and j where i < j, the query at position i becomes irrelevant because the query at position j will override it by bringing window 2 to the top again.

This observation leads us to process queries in reverse order. When traversing backwards:

• The first time we encounter a window in queries (which is actually its last query in the original order), we know this determines its final position
• Windows encountered earlier in reverse traversal should appear higher in the final result since they were queried later in the original sequence

By using a hash set s to track windows we've already placed, we can:

1. Add each uniquely queried window to our answer in the order we encounter them (reverse of queries)
2. Skip duplicate occurrences since we've already determined that window's final position
3. Finally, append any windows from the original arrangement that were never queried, maintaining their relative order

This approach elegantly handles the "most recent query wins" nature of the problem without actually simulating each individual window movement, achieving O(m + n) time complexity where m is the length of queries and n is the number of windows.

Solution Approach

The solution implements the reverse traversal strategy with a hash table to efficiently build the final window arrangement.

Step-by-step implementation:

1. Initialize data structures:

   • Create an empty set s to track windows that have already been placed in the result
   • Create an empty list ans to build the final window arrangement

2. Process queries in reverse order:

   for q in queries[::-1]:
       if q not in s:
           ans.append(q)
           s.add(q)

   • Use queries[::-1] to traverse the queries array from end to beginning
   • For each query q, check if it's already in the set s
   • If not present (first occurrence in reverse, which means last occurrence in original order):
     • Add q to the answer array ans
     • Add q to the set s to mark it as processed

3. Add remaining windows:

   for w in windows:
       if w not in s:
           ans.append(w)

   • Iterate through the original windows array
   • Add any window w that hasn't been queried (not in set s) to the answer
   • These windows maintain their relative order from the original arrangement

4. Return the result:

   • The ans array now contains the final window arrangement

Time Complexity: O(m + n) where m is the length of queries and n is the number of windows

• Reversing queries: O(m)
• Processing queries: O(m) with O(1) hash set operations
• Processing remaining windows: O(n)

Space Complexity: O(n) for the hash set storing at most n unique windows

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Given:

• windows = [3, 1, 4, 2] (initial order from top to bottom)
• queries = [4, 1, 4, 3] (sequence of windows to bring to top)

Step 1: Initialize data structures

• s = {} (empty set to track placed windows)
• ans = [] (empty list for final arrangement)

Step 2: Process queries in reverse order

We traverse queries backwards: [3, 4, 1, 4]

• Query 3: Not in s

  • Add 3 to ans: ans = [3]
  • Add 3 to s: s = {3}

• Query 4: Not in s

  • Add 4 to ans: ans = [3, 4]
  • Add 4 to s: s = {3, 4}

• Query 1: Not in s

  • Add 1 to ans: ans = [3, 4, 1]
  • Add 1 to s: s = {3, 4, 1}

• Query 4: Already in s (skip)

  • No changes to ans or s

Step 3: Add remaining windows from original arrangement

Check each window in windows = [3, 1, 4, 2]:

• Window 3: In s (skip)
• Window 1: In s (skip)
• Window 4: In s (skip)
• Window 2: Not in s
  • Add 2 to ans: ans = [3, 4, 1, 2]

Final Result: [3, 4, 1, 2]

Verification: Let's verify by simulating the original queries:

• Initial: [3, 1, 4, 2]
• Query 4: Move 4 to top → [4, 3, 1, 2]
• Query 1: Move 1 to top → [1, 4, 3, 2]
• Query 4: Move 4 to top → [4, 1, 3, 2]
• Query 3: Move 3 to top → [3, 4, 1, 2] ✓

The reverse processing correctly produces the same result without simulating each move!

Solution Implementation

Time and Space Complexity

The time complexity is O(n + m), where n is the length of the windows array and m is the length of the queries array.

Breaking down the time complexity:

• Reversing the queries array with queries[::-1] takes O(m) time
• The first loop iterates through all reversed queries, performing O(1) operations (set lookup, list append, set add) for each element, taking O(m) time total
• The second loop iterates through all windows, performing O(1) operations (set lookup, list append) for each element, taking O(n) time total
• Overall: O(m) + O(m) + O(n) = O(n + m)

The space complexity is O(m), where m is the length of the queries array.

Breaking down the space complexity:

• The set s stores at most m unique elements from queries (in the worst case, all queries are unique), requiring O(m) space
• The ans list stores at most n + m elements, but this is considered the output and typically not counted in space complexity analysis
• The reversed queries queries[::-1] creates a new list of size m, requiring O(m) space
• Overall auxiliary space: O(m)

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

Common Pitfalls

1. Processing Duplicate Queries Incorrectly

The Pitfall: A common mistake is not handling duplicate window IDs in the queries array properly. If you don't use a set to track already-processed windows, you might add the same window multiple times to the result.

Incorrect Implementation:

def simulationResult(self, windows: List[int], queries: List[int]) -> List[int]:
    result = []
  
    # Wrong: No tracking of processed windows
    for query_window in reversed(queries):
        result.append(query_window)  # Adds duplicates!
  
    for window in windows:
        if window not in result:  # O(n) operation each time!
            result.append(window)
  
    return result

Why it fails: If queries = [1, 2, 1], the reversed order [1, 2, 1] would add window 1 twice to the result, creating an invalid arrangement.

The Fix: Always use a set to track which windows have been processed:

processed_windows = set()
for query_window in reversed(queries):
    if query_window not in processed_windows:  # Check before adding
        result.append(query_window)
        processed_windows.add(query_window)

2. Using List Membership Check Instead of Set

The Pitfall: Using if window not in result instead of a separate set for tracking, which degrades performance.

Inefficient Implementation:

def simulationResult(self, windows: List[int], queries: List[int]) -> List[int]:
    result = []
  
    for query_window in reversed(queries):
        if query_window not in result:  # O(n) operation!
            result.append(query_window)
  
    for window in windows:
        if window not in result:  # O(n) operation!
            result.append(window)
  
    return result

Why it's problematic: List membership checking (in operator) has O(n) time complexity, making the overall solution O(n²) instead of O(n).

The Fix: Use a separate set for O(1) membership checking:

processed_windows = set()  # O(1) membership checking

3. Forgetting to Maintain Original Order for Non-Queried Windows

The Pitfall: Adding non-queried windows in the wrong order or using a data structure that doesn't preserve the original ordering.

Incorrect Implementation:

def simulationResult(self, windows: List[int], queries: List[int]) -> List[int]:
    result = []
    processed_windows = set()
  
    for query_window in reversed(queries):
        if query_window not in processed_windows:
            result.append(query_window)
            processed_windows.add(query_window)
  
    # Wrong: Using set difference loses original order!
    remaining = set(windows) - processed_windows
    result.extend(remaining)  # Order is not guaranteed
  
    return result

Why it fails: Sets don't maintain insertion order (in Python < 3.7) or the original order from the windows array, so the relative positions of non-queried windows would be lost.

The Fix: Iterate through the original windows array to preserve order:

for window in windows:  # Maintains original order
    if window not in processed_windows:
        result.append(window)