LeetCode 2293: Min Max Game Solution

Problem Explanation

Explanation:

The given problem requires repeatedly applying a specific algorithm on an array until only one number remains.
At each step, we create a new array by taking the minimum of pair values at even indices and the maximum of pair values at odd indices from the current array.
We continue this process until the length of the array becomes 1, and then return the remaining number.

Algorithmic Idea:

Start with the given array.
While the length of the array is greater than 1, create a new array by taking the min/max values at even/odd indices.
Replace the current array with the new array.
Repeat steps 2 and 3 until the length of the array becomes 1.
Return the remaining number.

Time Complexity: O(n log n) where n is the length of the input array nums.

Space Complexity: O(n) for creating new arrays at each step.

Solution Code

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

Frequently Asked Questions

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What is the time complexity of LeetCode 2293 (Min Max Game)?

The time complexity for LeetCode 2293 (Min Max Game) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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2293. Min Max Game