How do you solve Min Max Game on LeetCode?

Min Max Game (LeetCode #2293) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The algorithm reduces the size of the array exponentially, leading to a logarithmic number of iterations.

What is the brute force solution for Min Max Game?

The brute force approach for Min Max Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates the entire algorithm step by step. We repeatedly create new arrays by applying the min and max operations until only one number remains. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Min Max Game?

Min Max Game uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Simulation.

What are the interview tips for Min Max Game?

Always clarify the constraints and edge cases before starting to code. Think about how you can optimize your solution after implementing the brute force version. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Min Max Game?

Not recognizing that the length of the array is always a power of 2, which simplifies the logic. Failing to handle the case where the array length is 1 upfront.

#2293

Min Max Game

Easy

ArraySimulationArraySimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses the same simulation approach but recognizes that we can directly compute the last remaining number without storing intermediate arrays, reducing space usage.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to hold the current index.
2Step 2: Iterate through the nums array, applying min/max operations directly in a single pass.
3Step 3: Keep reducing the size of the array until only one element remains.
4Step 4: Return the last remaining number.

solution.py11 lines

1# Full working Python code
2
3def minMaxGame(nums):
4    while len(nums) > 1:
5        for i in range(len(nums) // 2):
6            if i % 2 == 0:
7                nums[i] = min(nums[2 * i], nums[2 * i + 1])
8            else:
9                nums[i] = max(nums[2 * i], nums[2 * i + 1])
10        nums = nums[:len(nums) // 2]
11    return nums[0]

ℹ

Complexity note: This complexity is linear because we are processing each element of the array a constant number of times, and we do not use any extra space proportional to the input size.

1The algorithm reduces the size of the array exponentially, leading to a logarithmic number of iterations.
2Understanding the pattern of min/max operations helps in optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.