#2293

Easy

LeetCode Problem Workspace

Min Max Game

The Min Max Game problem requires simulating an array reduction process to find the last remaining number.

Problem Statement

You are given a 0-indexed integer array nums, where its length is a power of 2. The task is to simulate a reduction process on the array, applying a specific operation repeatedly until only one number remains.

In each step of the reduction, you consider pairs of adjacent elements: at even indices, the minimum of the pair is kept, and at odd indices, the maximum is kept. This process continues until there is only one element left, which is the answer.

Examples

Example 1

Input: nums = [1,3,5,2,4,8,2,2]

Output: 1

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

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.

Solution Approach

Simulate the Process

Start by simulating the pairwise operation on the array, applying the minimum for even indices and the maximum for odd indices, until the array has only one element left.

Optimizing with Iterative Reductions

Iterate through the array, shrinking it by half each time while performing the necessary operations. This approach ensures the problem is solved efficiently within the given constraints.

Handle Edge Cases

Ensure proper handling of edge cases like when the array has only one element or when all elements are equal, ensuring correctness for all possible inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the number of elements in the array, because the array is halved in size each time. The space complexity is O(1), since the operation modifies the array in place without requiring additional space.

What Interviewers Usually Probe

The candidate correctly simulates the process without overcomplicating the solution.
The candidate handles edge cases and large inputs efficiently.
The candidate explains how the time complexity is reduced with each step of the iteration.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the problem and applying the wrong operation for even and odd indices.
Not recognizing that the length of the array is halved at each step, which can lead to inefficient solutions.
Failing to handle edge cases such as when the array has only one element or when all elements are equal.

Follow-up variants

Simulate with different operations (min vs max), alternating by indices or positions.
Modify the problem to find the minimum/maximum value of the remaining number, instead of just the last one.
Apply the same process but with different array lengths (not restricted to power of 2).

FAQ

What is the Min Max Game problem?

The Min Max Game problem asks you to simulate a reduction process on an array where you apply pairwise minimum or maximum operations until one number remains.

How can I solve the Min Max Game problem efficiently?

To solve it efficiently, simulate the reduction process iteratively, reducing the array by half in each step while applying the minimum for even indices and the maximum for odd indices.

What is the time complexity of the Min Max Game problem?

The time complexity is O(n), where n is the length of the array, because the size of the array is halved with each reduction step.

Are there any edge cases to consider in the Min Max Game problem?

Yes, edge cases include when the array has only one element or when all elements in the array are equal.

What patterns does the Min Max Game problem focus on?

The problem focuses on array simulation and the efficient application of pairwise operations to reduce the array.

terminal

Solution

Solution 1: Simulation

According to the problem statement, we can simulate the entire process, and the remaining number will be the answer. In implementation, we do not need to create an additional array; we can directly operate on the original array.

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class Solution:
    def minMaxGame(self, nums: List[int]) -> int:
        n = len(nums)
        while n > 1:
            n >>= 1
            for i in range(n):
                a, b = nums[i << 1], nums[i << 1 | 1]
                nums[i] = min(a, b) if i % 2 == 0 else max(a, b)
        return nums[0]

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