LeetCode in Kotlin

2293. Min Max Game

Easy

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

Apply the following algorithm on nums:

1. 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.
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]).
3. 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]).
4. Replace the array nums with newNums.
5. 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

class Solution {
    fun minMaxGame(nums: IntArray): Int {
        val n = nums.size
        if (n == 1) {
            return nums[0]
        }
        val newNums = IntArray(n / 2)
        for (i in 0 until n / 2) {
            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)
    }
}

This site is open source. Improve this page.