2293. Min Max Game / TypeScript

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.

 

출처 : https://leetcode.com/problems/min-max-game/description/

 

 

풀이 

function minMaxGame(nums: number[]): number {
    if (nums.length === 1) return nums[0];
    if (nums.length === 2) return Math.min(nums[0], nums[1]);
    
    const newNums = [];
    for (let i = 0; i < nums.length; i+=4) {
        newNums.push(Math.min(nums[i], nums[i+1]));
        newNums.push(Math.max(nums[i+2], nums[i+3]));
    }
    
    return minMaxGame(newNums);
};