2293

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.

Related Topics:
Array, Simulation

Similar Questions:

• Elimination Game (Medium)
• Find Triangular Sum of an Array (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/min-max-game
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int minMaxGame(vector<int>& A) {
        while (A.size() > 1) {
            vector<int> next(A.size() / 2);
            for (int i = 0; i < next.size(); ++i) {
                if (i % 2 == 0) next[i] = min(A[2 * i], A[2 * i + 1]);
                else next[i] = max(A[2 * i], A[2 * i + 1]);
            }
            swap(next, A);
        }
        return A[0];
    }
};

Or compute in place

// OJ: https://leetcode.com/problems/min-max-game
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int minMaxGame(vector<int>& A) {
        int N = A.size();
        while (N > 1) {
            N /= 2;
            for (int i = 0; i < N; ++i) {
                if (i % 2 == 0) A[i] = min(A[2 * i], A[2 * i + 1]);
                else A[i] = max(A[2 * i], A[2 * i + 1]);
            }
        }
        return A[0];
    }
};