Leetcode 2293 Min Max Game Solution in c++ | Hindi Coding Community

 

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

Apply the following algorithm on nums:

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.

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]).

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]).

Replace the array nums with newNums.

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.

int minMaxGame(vector<int>& a) {

for(int n= size(a); n>1 ; n-=(n/2)){

for(int i=0;i<n/2;i++)

a[i]= (i%2)==1? max(a[2 * i], a[2 * i + 1]) : min(a[2 * i], a[2 * i + 1]);

}

return a[0];

}