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Formatted question description: https://leetcode.ca/all/2293.html

2293. Min Max Game

  • Difficulty: Easy.
  • Related Topics: Array, Simulation.
  • Similar Questions: Elimination Game, Find Triangular Sum of an Array.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public int minMaxGame(int[] nums) {
        int n = nums.length;
        if (n == 1) {
            return nums[0];
        }
        int[] newNums = new int[n / 2];
        for (int i = 0; i < n / 2; i++) {
            if (i % 2 == 0) {
                newNums[i] = Math.min(nums[2 * i], nums[2 * i + 1]);
            } else {
                newNums[i] = Math.max(nums[2 * i], nums[2 * i + 1]);
            }
        }
        return minMaxGame(newNums);
    }
}

  • Todo

  • 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]

############

# 2293. Min Max Game
# https://leetcode.com/problems/min-max-game/

class Solution:
    def minMaxGame(self, nums: List[int]) -> int:
        
        while len(nums) != 1:
            temp = []
            
            for i in range(len(nums) // 2):
                if i % 2 == 0:
                    temp.append(min(nums[i * 2], nums[i * 2 + 1]))
                else:
                    temp.append(max(nums[i * 2], nums[i * 2 + 1]))
            
            nums = temp
        
        return nums[0]

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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