3194. Minimum Average of Smallest and Largest Elements

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3194. Minimum Average of Smallest and Largest Elements

Description

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:

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Input: nums = [7,8,3,4,15,13,4,1]

Output: 5.5

Explanation:

stepnumsaverages
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:

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Input: nums = [1,9,8,3,10,5]

Output: 5.5

Explanation:

stepnumsaverages
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:

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Input: nums = [1,2,3,7,8,9]

Output: 5.0

Explanation:

stepnumsaverages
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

Hints/Notes

  • Weekly Contest 403

Solution

Language: C++

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class Solution {
public:
    double minimumAverage(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int left = 0, right = nums.size() - 1;
        double res = INT_MAX;
        while (left < right) {
            res = min((nums[left] + nums[right]) * 1.0 / 2, res);
            left++;
            right--;
        }
        return res;
    }
};