3194_Minimum_Average_of_Smallest_and_Largest_Elements.cpp

/*
3194. Minimum Average of Smallest and Largest Elements
You have an array of floating point numbers averages which is initially empty. You are given an array nums of n integers where n is even.
You repeat the following procedure n / 2 times:
Remove the smallest element, minElement, and the largest element maxElement, from nums.
Add (minElement + maxElement) / 2 to averages.
Return the minimum element in averages.
Example 1:
Input: nums = [7,8,3,4,15,13,4,1]
Output: 5.5
Explanation:
step	nums	averages
0	[7,8,3,4,15,13,4,1]	[]
1	[7,8,3,4,13,4]	[8]
2	[7,8,4,4]	[8,8]
3	[7,4]	[8,8,6]
4	[]	[8,8,6,5.5]
The smallest element of averages, 5.5, is returned.
Example 2:
Input: nums = [1,9,8,3,10,5]
Output: 5.5
Explanation:
step	nums	averages
0	[1,9,8,3,10,5]	[]
1	[9,8,3,5]	[5.5]
2	[8,5]	[5.5,6]
3	[]	[5.5,6,6.5]
Example 3:
Input: nums = [1,2,3,7,8,9]
Output: 5.0
Explanation:
step	nums	averages
0	[1,2,3,7,8,9]	[]
1	[2,3,7,8]	[5]
2	[3,7]	[5,5]
3	[]	[5,5,5]

Constraints:
2 <= n == nums.length <= 50
n is even.
1 <= nums[i] <= 50
*/
/*
    TWO POINTER APPROACH
Time Complexity O(n*Log(n))
Space Complexity O(1)
*/
class Solution {
public:
    double minimumAverage(vector<int>& nums) {
        double avg = 100 ;
     sort(nums.begin(),nums.end());   
        int i = 0 , j = nums.size()-1 , count =0 ;
        while(i<=j && count<nums.size()/2){
            avg = min(avg,(double)(nums[i]+nums[j])/2);
            i++;j--;count++;
        }
        return avg;
    }
};