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: Prefix / Suffix Sum
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int minimumAverageDifference(vector<int>& nums) { const int n = nums.size(); long long suffix = accumulate(begin(nums), end(nums), 0LL); long long prefix = 0; int best = INT_MAX; int ans = 0; for (int i = 0; i < n; ++i) { prefix += nums[i]; suffix -= nums[i]; int cur = abs(prefix / (i + 1) - (i == n - 1 ? 0 : suffix / (n - i - 1))); if (cur < best) { best = cur; ans = i; } } return ans; } }; |
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