Min Max Game

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ArraysRecursion

You are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm on nums:

Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
Replace the array nums with newNums.
Repeat the entire process starting from step 1.

Return the last number that remains in nums after applying the algorithm.

Example 1:

Input: nums = [1,3,5,2,4,8,2,2]
Output: 1
Explanation: The following arrays are the results of applying the algorithm repeatedly.
First: nums = [1,5,4,2]
Second: nums = [1,4]
Third: nums = [1]
1 is the last remaining number, so we return 1.

Example 2:

Input: nums = [3]
Output: 3
Explanation: 3 is already the last remaining number, so we return 3.

Constraints:

1 <= nums.length <= 1024
1 <= nums[i] <= 10⁹
nums.length is a power of 2.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

To confirm the base case, if the input array initially contains only one element, should the function return that element immediately?
The problem description mentions replacing `nums` with `newNums`. Is it acceptable to allocate a new array for each step of the reduction, or is an in-place modification of the array preferred for better space complexity?
The constraints state that `nums.length` is a power of 2. Can I safely assume this will always be true, or should I add validation for inputs that do not meet this criterion?
The problem defines the operations based on the parity of the index `i` in the *new* array. For instance, `newNums[0]` would be a `min` operation and `newNums[1]` a `max` operation. Is this understanding correct?
The constraints mention that array elements are positive integers. Just to be thorough, do I need to consider the possibility of handling negative numbers or zeros in the input array?

Brute Force Solution

Approach

The brute force strategy is to directly simulate the game round by round, exactly as the rules dictate. We'll start with the full list of numbers and repeatedly shrink it by applying the min/max rule to pairs until only one number is left.

Here's how the algorithm would work step-by-step:

Start with the initial list of numbers given.
As long as the list contains more than one number, you need to perform another round.
To start a round, prepare a new, empty list which will be half the size of the current one.
Look at the numbers in the current list as consecutive pairs, from left to right.
For the first pair, find the smaller number and place it in your new list.
For the second pair, find the larger number and place it in your new list.
Continue this alternating pattern for all the remaining pairs: take the smaller value from the third pair, the larger from the fourth, and so on.
Once you have processed all the pairs, replace your original list with this newly created shorter list.
Repeat this entire process until your list has been reduced to a single number.
This final number is the result of the game.

Code Implementation

class Solution:
    def minMaxGame(self, nums: list[int]) -> int:
        current_numbers = list(nums)

        # The game continues as long as there is more than one number left to process.
        while len(current_numbers) > 1:

            list_size = len(current_numbers)
            next_round_numbers = []

            # Iterate through the current numbers in pairs to generate the next list for the subsequent round.
            for pair_index in range(list_size // 2):
                first_element = current_numbers[2 * pair_index]
                second_element = current_numbers[2 * pair_index + 1]

                # Alternate between min and max operations based on the pair's position (even or odd index).
                if pair_index % 2 == 0:
                    next_round_numbers.append(min(first_element, second_element))
                else:
                    next_round_numbers.append(max(first_element, second_element))
            
            # The newly generated list becomes the list for the next iteration of the game.
            current_numbers = next_round_numbers
        
        return current_numbers[0]

Big(O) Analysis

Time Complexity

O(n) – The time complexity is determined by the total number of operations performed across all rounds of the simulation. The process starts with a list of `n` elements and repeatedly halves the list size. The first round processes `n/2` pairs, the second round processes `n/4` pairs, and this continues until only one element remains. The total number of operations is the sum of a geometric series, `n/2 + n/4 + n/8 + ... + 1`, which totals `n-1`. Since the total work is directly proportional to the initial input size, the complexity simplifies to O(n).

Space Complexity

O(N) – The algorithm's space usage is determined by the 'new, empty list' created in each round of the simulation. In the first round, given an input list of size N, a new list of size N/2 is allocated to store the intermediate results. This is the peak auxiliary memory usage, as the lists created in subsequent rounds are progressively smaller. Therefore, the extra space required is directly proportional to the initial input size N, leading to a space complexity of O(N).

Optimal Solution

Approach

The optimal strategy is a direct simulation of the game. You simply follow the rules round by round, progressively shrinking the list of numbers until only one remains.

Here's how the algorithm would work step-by-step:

Start with the initial list of numbers.
As long as the list contains more than one number, you will perform a reduction step.
In each step, you'll create a new, temporary list that is half the size of the current one.
Process the current list by taking adjacent pairs of numbers, starting from the beginning.
For each pair, decide whether to take the smaller or larger number based on the spot you are filling in the new list.
If you are filling an even-numbered spot, like the first or third, add the smaller of the two numbers to the new list.
If you are filling an odd-numbered spot, like the second or fourth, add the larger of the two numbers to the new list.
After converting all pairs into new numbers, replace the old list with this new, shorter one.
Continue this reduction process until only one number is left. That single number is the final answer.

Code Implementation

def min_max_game(numbers_list):
    current_numbers = list(numbers_list)

    # The simulation proceeds in rounds, shrinking the list until only one number remains.
    while len(current_numbers) > 1:
        next_round_length = len(current_numbers) // 2
        next_round_numbers = [0] * next_round_length

        # Each new element is derived from a pair in the current list based on an alternating rule.
        for new_list_index in range(next_round_length):
            # The rule depends on the new element's index: min for even, max for odd.
            if new_list_index % 2 == 0:
                first_num_in_pair = current_numbers[2 * new_list_index]
                second_num_in_pair = current_numbers[2 * new_list_index + 1]
                next_round_numbers[new_list_index] = min(first_num_in_pair, second_num_in_pair)
            else:
                first_num_in_pair = current_numbers[2 * new_list_index]
                second_num_in_pair = current_numbers[2 * new_list_index + 1]
                next_round_numbers[new_list_index] = max(first_num_in_pair, second_num_in_pair)
        
        current_numbers = next_round_numbers

    return current_numbers[0]

Big(O) Analysis

Time Complexity

O(n) – The algorithm's runtime is determined by the total number of comparisons across all rounds. Starting with a list of size n, the first round involves n/2 comparisons to create a new list of size n/2. The next round performs n/4 comparisons, and this process of halving continues until one number is left. The total number of operations is the sum of a geometric series, n/2 + n/4 + n/8 + ... + 1. This sum is exactly n-1, which simplifies to a linear time complexity of O(n).

Space Complexity

O(N) – The dominant factor in space complexity is the creation of a "new, temporary list" in each reduction step as described in the simulation. This temporary list is used to hold the intermediate results of the current round. The largest this list will be is during the first step, where it needs to hold N/2 elements, with N being the size of the initial input list. Since the memory required for this list is directly proportional to the input size N, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Input array has only one element	The algorithm's base case is n=1, so the loop does not run and the single element is returned immediately.
Input array has the smallest non-trivial size of two elements	The process runs once, calculating the minimum of the two elements since the new index is 0 (even).
Input array has the maximum allowed length (1024)	The iterative reduction is efficient, as the total operations are proportional to n + n/2 + n/4..., which is O(n).
All elements in the input array are identical	Since min(x, x) and max(x, x) both equal x, the algorithm correctly propagates the identical value to the final result.
An input of length 4, the smallest case testing both min and max rules	The first reduction step must correctly apply min() for the even index 0 and max() for the odd index 1.
Input contains values at the boundaries of the constraints, like 1 and 10^9	The standard integer types and min/max functions handle this range correctly without any risk of overflow.
The problem states the input length is a power of 2	The solution relies on this guarantee, as it ensures the array can be perfectly halved until a single element remains.
Implementing the solution with in-place array modification	This optimization is safe because the write index `i` is always smaller than the read indices `2i` and `2i+1`.

Input array has only one element

How to Handle:

The algorithm's base case is n=1, so the loop does not run and the single element is returned immediately.

Input array has the smallest non-trivial size of two elements

How to Handle:

The process runs once, calculating the minimum of the two elements since the new index is 0 (even).

Input array has the maximum allowed length (1024)

How to Handle:

The iterative reduction is efficient, as the total operations are proportional to n + n/2 + n/4..., which is O(n).

All elements in the input array are identical

How to Handle:

Since min(x, x) and max(x, x) both equal x, the algorithm correctly propagates the identical value to the final result.

An input of length 4, the smallest case testing both min and max rules

How to Handle:

The first reduction step must correctly apply min() for the even index 0 and max() for the odd index 1.

Input contains values at the boundaries of the constraints, like 1 and 10^9

How to Handle:

The standard integer types and min/max functions handle this range correctly without any risk of overflow.

The problem states the input length is a power of 2

How to Handle:

The solution relies on this guarantee, as it ensures the array can be perfectly halved until a single element remains.

Implementing the solution with in-place array modification

How to Handle:

This optimization is safe because the write index `i` is always smaller than the read indices `2*i` and `2*i+1`.