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:
- The absolute difference of two numbers is the absolute value of their difference.
- The average of
n
elements is the sum of then
elements divided (integer division) byn
. - The average of
0
elements is considered to be0
.
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 <= 105
-
0 <= nums[i] <= 105
SOLUTION:
class Solution:
def getDiff(self, prefix, i, n):
l = (prefix[i + 1] - prefix[0])
r = prefix[-1] - l
l = l // (i + 1)
r = r // (n - i - 1) if i < n - 1 else 0
return abs(l - r)
def minimumAverageDifference(self, nums: List[int]) -> int:
n = len(nums)
prefix = [0]
for i in range(n):
prefix.append(prefix[-1] + nums[i])
return min(range(n), key = lambda i: self.getDiff(prefix, i, n))
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