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2293. Min Max Game

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

  • class Solution {
    public int minMaxGame(int[] nums) {
        for (int n = nums.length; n > 1;) {
            n >>= 1;
            for (int i = 0; i < n; ++i) {
                int a = nums[i << 1], b = nums[i << 1 | 1];
                nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
            }
        }
        return nums[0];
    }
}

  • class Solution {
public:
    int minMaxGame(vector<int>& nums) {
        for (int n = nums.size(); n > 1;) {
            n >>= 1;
            for (int i = 0; i < n; ++i) {
                int a = nums[i << 1], b = nums[i << 1 | 1];
                nums[i] = i % 2 == 0 ? min(a, b) : max(a, b);
            }
        }
        return nums[0];
    }
};

  • 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]


  • func minMaxGame(nums []int) int {
	for n := len(nums); n > 1; {
		n >>= 1
		for i := 0; i < n; i++ {
			a, b := nums[i<<1], nums[i<<1|1]
			if i%2 == 0 {
				nums[i] = min(a, b)
			} else {
				nums[i] = max(a, b)
			}
		}
	}
	return nums[0]
}

  • function minMaxGame(nums: number[]): number {
    for (let n = nums.length; n > 1; ) {
        n >>= 1;
        for (let i = 0; i < n; ++i) {
            const a = nums[i << 1];
            const b = nums[(i << 1) | 1];
            nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
        }
    }
    return nums[0];
}


  • impl Solution {
    pub fn min_max_game(mut nums: Vec<i32>) -> i32 {
        let mut n = nums.len();
        while n != 1 {
            n >>= 1;
            for i in 0..n {
                nums[i] = (if (i & 1) == 1 { i32::max } else { i32::min })(
                    nums[i << 1],
                    nums[(i << 1) | 1]
                );
            }
        }
        nums[0]
    }
}

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