You are given a 0-indexed integer array nums of length n.
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.
Return the index with the minimum average difference. If there are multiple such indices, return the smallest one.
Note:
n elements is the sum of the n elements divided (integer division) by n.0 elements is considered to be 0.
Example 1:
Input: nums = [2,5,3,9,5,3] Output: 3 Explanation: - The average difference of index 0 is: |2 / 1 - (5 + 3 + 9 + 5 + 3) / 5| = |2 / 1 - 25 / 5| = |2 - 5| = 3. - The average difference of index 1 is: |(2 + 5) / 2 - (3 + 9 + 5 + 3) / 4| = |7 / 2 - 20 / 4| = |3 - 5| = 2. - The average difference of index 2 is: |(2 + 5 + 3) / 3 - (9 + 5 + 3) / 3| = |10 / 3 - 17 / 3| = |3 - 5| = 2. - The average difference of index 3 is: |(2 + 5 + 3 + 9) / 4 - (5 + 3) / 2| = |19 / 4 - 8 / 2| = |4 - 4| = 0. - The average difference of index 4 is: |(2 + 5 + 3 + 9 + 5) / 5 - 3 / 1| = |24 / 5 - 3 / 1| = |4 - 3| = 1. - The average difference of index 5 is: |(2 + 5 + 3 + 9 + 5 + 3) / 6 - 0| = |27 / 6 - 0| = |4 - 0| = 4. The average difference of index 3 is the minimum average difference so return 3.
Example 2:
Input: nums = [0] Output: 0 Explanation: The only index is 0 so return 0. The average difference of index 0 is: |0 / 1 - 0| = |0 - 0| = 0.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 105class Solution {
public:
int minimumAverageDifference(vector<int>& n) {
vector <long> nums = {n[0]};
for(int i=1; i<n.size(); i++)
nums.push_back(n[i]+nums.back());
int diff = INT_MAX, sum=nums.back(), len=nums.size(), res=0;
for(int i=0; i<len; i++){
int avgLt = round(nums[i]/(i+1));
int avgRt = round((nums.back()-nums[i])/max(len-i-1, 1));
int cur = abs(avgLt-avgRt);
if(diff>cur)
diff = cur, res = i;
// cout << avgLt << "\t" << avgRt << endl;
}
return res;
}
};