Minimum Average Difference

정수들이 담긴 벡터를 받고,

벡터에서 특정 두 구간에서의 평균 차이가 가장 작을 때 인덱스를 구하는 문제

두 구간의 조건은 아래와 같다

The average difference of the index i is the absolute difference between the average of the first i + 1 elements of nums and the average of the last n - i - 1 elements. Both averages should be rounded down to the nearest integer.

class Solution {
public:
    int minimumAverageDifference(vector<int>& nums) {
        
        int numsSize = nums.size();
        vector<long long int> sums(numsSize);

        sums[0] = nums[0];
        for (int i = 1; i < numsSize; i++)
        {
            sums[i] = sums[i - 1] + nums[i];
        }

        long long int min{numeric_limits<int>::max()};
        int minIndex{0};
        for (int i = 0; i < numsSize; i++)
        {
            int head{i + 1};
            int tail{numsSize - head};

            long long int headMean = sums[i];
            long long int tailMean = sums[numsSize - 1] - sums[i];

            int temp{};
            if (tail == 0)
            {
                temp = abs(headMean / head);
            }
            else
            {
                temp = abs(floor(headMean / head) - floor(tailMean / tail));
            }

            if (temp < min)
            {
                min = temp;
                minIndex = i;
            }
        }

        return minIndex;
    }
};