Problem Description

You have an integer array nums with an even number of elements (0-indexed) and an empty array arr. Alice and Bob are playing a game with these arrays.

The game follows these rules for each round:

1. First, Alice removes the minimum element from nums
2. Then, Bob removes the (new) minimum element from nums
3. Next, Bob adds his removed element to arr
4. Finally, Alice adds her removed element to arr

This process repeats until nums is empty. Your task is to return the final arr array.

Notice the key twist: while Alice picks first, Bob gets to add his element to the result array first. This means if Alice picks element a and Bob picks element b in a round, they get added to arr in the order [b, a].

For example, if nums = [5, 4, 2, 3]:

• Round 1: Alice takes 2, Bob takes 3 → arr becomes [3, 2]
• Round 2: Alice takes 4, Bob takes 5 → arr becomes [3, 2, 5, 4]

The solution uses a min heap to efficiently extract the minimum elements. In each iteration, it pops two minimums a and b (where a ≤ b), then appends them in reversed order [b, a] to match the game rules.

Intuition

The game has a clear pattern: in each round, we need to find and remove the two smallest elements from the remaining array. This immediately suggests using a data structure that efficiently gives us the minimum element - a min heap.

Let's trace through what happens in each round:

• Alice removes the smallest element (let's call it a)
• Bob removes the next smallest element (let's call it b)
• They add these to arr in the order [b, a] (Bob first, then Alice)

Since we always need the two smallest elements in each round, and after removing them we might expose new minimums for the next round, a min heap is perfect. It maintains the minimum element at the root and automatically reorganizes after each removal.

The key insight is that we can simplify the game mechanics. Instead of simulating Alice and Bob's turns separately, we can:

1. Extract two minimums at once (since we know both players will take the smallest available)
2. Add them to the result in reversed order (since Bob adds first despite picking second)

This transforms the problem from a complex turn-based simulation into a simple pattern: repeatedly pop two minimums from a heap and append them in reversed order. The heapify operation converts our array into a min heap in O(n) time, and each heappop takes O(log n) time, making this approach efficient.

The beauty of this solution is recognizing that the game's complexity is just an illusion - it's really just about pairing up elements in sorted order and swapping each pair.

Learn more about Sorting and Heap (Priority Queue) patterns.

Solution Approach

The solution uses a min heap (priority queue) to efficiently extract minimum elements. Here's the step-by-step implementation:

1. Convert array to min heap:

heapify(nums)

This transforms the input array nums into a min heap in-place in O(n) time. After this operation, nums[0] will always be the smallest element.

2. Initialize result array:

ans = []

Create an empty list to store the final result.

3. Process pairs of elements:

while nums:
    a, b = heappop(nums), heappop(nums)
    ans.append(b)
    ans.append(a)

The main loop continues until the heap is empty:

• Extract the smallest element and store it in a (Alice's pick)
• Extract the next smallest element and store it in b (Bob's pick)
• Append b first (Bob adds his element first)
• Append a second (Alice adds her element second)

Each heappop() operation:

• Removes and returns the minimum element from the heap
• Automatically restructures the heap to maintain the min heap property
• Takes O(log n) time

Time Complexity: O(n log n) where n is the length of the input array. We perform n heap pop operations, each taking O(log n) time.

Space Complexity: O(1) for the heap operations (since we modify in-place), but O(n) for the result array that we need to return.

The elegance of this approach lies in recognizing that we don't need to simulate the game turn by turn. Instead, we can process each round (two picks) at once and directly construct the result in the correct order.

Example Walkthrough

Let's trace through the solution with nums = [2, 1, 3, 4]:

Step 1: Convert to min heap After heapify(nums), the array becomes a min heap structure. Internally it might look like [1, 2, 3, 4] (the exact internal structure may vary but maintains heap property where parent ≤ children).

Step 2: Initialize result ans = []

Step 3: Process pairs

Round 1:

• First heappop(nums) returns 1 (smallest element) → a = 1 (Alice's pick)
• Heap now contains [2, 4, 3] (restructured after removal)
• Second heappop(nums) returns 2 (new smallest) → b = 2 (Bob's pick)
• Heap now contains [3, 4]
• Append b then a: ans = [2, 1]

Round 2:

• First heappop(nums) returns 3 → a = 3 (Alice's pick)
• Heap now contains [4]
• Second heappop(nums) returns 4 → b = 4 (Bob's pick)
• Heap is now empty
• Append b then a: ans = [2, 1, 4, 3]

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

Notice how in each round:

• We extract the two smallest remaining elements
• The second element extracted (Bob's) goes into the result first
• This creates the pattern where each pair is reversed from their extraction order

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log n), where n is the length of the array nums. This is because:

• The heapify() operation takes O(n) time to build a min-heap from the array
• The while loop runs n/2 iterations (processing two elements per iteration)
• Each heappop() operation takes O(log n) time to maintain the heap property
• Total: O(n) + n × O(log n) = O(n × log n)

The space complexity is O(n), where n is the length of the array nums. This is because:

• The ans list stores all n elements from the original array, requiring O(n) space
• The heapify operation is performed in-place on the input array, so no additional space is needed for the heap itself
• Overall space usage is dominated by the output array ans

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

Common Pitfalls

1. Forgetting to handle the reversed order of insertion

A common mistake is to append elements in the order they were extracted rather than reversing them:

Incorrect:

while nums:
    a = heappop(nums)
    b = heappop(nums)
    result.append(a)  # Wrong order!
    result.append(b)

Correct:

while nums:
    a = heappop(nums)
    b = heappop(nums)
    result.append(b)  # Bob's element first
    result.append(a)  # Alice's element second

2. Not validating input assumptions

The problem states the array has an even number of elements, but production code should handle edge cases:

Vulnerable:

def numberGame(self, nums: List[int]) -> List[int]:
    heapify(nums)
    result = []
    while nums:  # Will crash if odd number of elements
        a = heappop(nums)
        b = heappop(nums)  # IndexError if nums is empty!

Robust:

def numberGame(self, nums: List[int]) -> List[int]:
    if len(nums) % 2 != 0:
        raise ValueError("Input must have even number of elements")
  
    heapify(nums)
    result = []
    while len(nums) >= 2:
        a = heappop(nums)
        b = heappop(nums)
        result.append(b)
        result.append(a)
    return result

3. Using sorted() instead of heap for simplicity but losing efficiency

While sorting the entire array upfront seems simpler, it's less efficient for very large arrays if we only need to process elements incrementally:

Less efficient for streaming scenarios:

def numberGame(self, nums: List[int]) -> List[int]:
    nums.sort()  # O(n log n) upfront cost
    result = []
    for i in range(0, len(nums), 2):
        result.append(nums[i+1])
        result.append(nums[i])
    return result

While this works and has the same overall complexity, the heap approach is better if the input could be streamed or if we might need to stop early in some variations of the problem.

4. Modifying the original array without considering side effects

The heapify() operation modifies the input array in-place. If the caller needs the original array preserved:

Problem:

def numberGame(self, nums: List[int]) -> List[int]:
    heapify(nums)  # Original nums array is modified!
    # ...

Solution:

def numberGame(self, nums: List[int]) -> List[int]:
    nums_copy = nums.copy()  # Create a copy first
    heapify(nums_copy)
    result = []
    while nums_copy:
        a = heappop(nums_copy)
        b = heappop(nums_copy)
        result.append(b)
        result.append(a)
    return result