README_EN.md

2293. Min Max Game

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Description

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.

Solutions

Python3

class Solution:
    def minMaxGame(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 1:
            return nums[0]
        t = []
        for i in range(n >> 1):
            v = max(nums[i << 1], nums[i << 1 | 1]) if i & 1 else min(
                nums[i << 1], nums[i << 1 | 1])
            t.append(v)
        return self.minMaxGame(t)

Java

class Solution {
    public int minMaxGame(int[] nums) {
        int n = nums.length;
        if (n == 1) {
            return nums[0];
        }
        int[] t = new int[n >> 1];
        for (int i = 0; i < t.length; ++i) {
            int a = nums[i << 1], b = nums[i << 1 | 1];
            t[i] = (i & 1) == 1 ? Math.max(a, b) : Math.min(a, b);
        }
        return minMaxGame(t);
    }
}

C++

class Solution {
public:
    int minMaxGame(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return nums[0];
        vector<int> t(n >> 1);
        for (int i = 0; i < t.size(); ++i)
        {
            int a = nums[i << 1], b = nums[i << 1 | 1];
            t[i] = (i & 1) == 1 ? max(a, b) : min(a, b);
        }
        return minMaxGame(t);
    }
};

Go

func minMaxGame(nums []int) int {
	n := len(nums)
	if n == 1 {
		return nums[0]
	}
	var t []int
	for i := 0; i < n>>1; i++ {
		a, b := nums[i<<1], nums[i<<1|1]
		if (i & 1) == 1 {
			t = append(t, max(a, b))
		} else {
			t = append(t, min(a, b))
		}
	}
	return minMaxGame(t)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

TypeScript

function minMaxGame(nums: number[]): number {
    while(nums.length > 1) {
        let n = nums.length;
        let tmp = [];
        for (let i = 0; i < n; i += 2) {
            if (i % 4 == 2) {
                tmp.push(Math.max(nums[i], nums[i + 1]));
            } else {
                tmp.push(Math.min(nums[i], nums[i + 1]));
            }
        }
        nums = tmp;
    }
    return nums[0];
};

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