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Formatted question description: https://leetcode.ca/all/2256.html

# 2256. Minimum Average Difference

• Difficulty: Medium.
• Related Topics: Array, Prefix Sum.
• Similar Questions: Split Array With Same Average, Number of Ways to Split Array.

## Problem

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 the n elements divided (integer division) by n.

• The average of 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 = 
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 (Java, C++, Python)

• class Solution {
public int minimumAverageDifference(int[] nums) {
long numsSum = 0;
for (int num : nums) {
numsSum += num;
}
long minAverageDiff = Long.MAX_VALUE;
long sumFromFront = 0;
int index = 0;
for (int i = 0; i < nums.length; i++) {
sumFromFront += nums[i];
int numbersRight = i == nums.length - 1 ? 1 : nums.length - i - 1;
long averageDiff =
Math.abs(sumFromFront / (i + 1) - (numsSum - sumFromFront) / numbersRight);
if (minAverageDiff > averageDiff) {
minAverageDiff = averageDiff;
index = i;
}
if (averageDiff == 0) {
break;
}
}
return index;
}
}

• Todo

• class Solution:
def minimumAverageDifference(self, nums: List[int]) -> int:
s = list(accumulate(nums))
ans, n = 0, len(nums)
mi = inf
for i in range(n):
a = s[i] // (i + 1)
b = 0 if i == n - 1 else (s[-1] - s[i]) // (n - i - 1)
t = abs(a - b)
if mi > t:
ans = i
mi = t
return ans

############

# 2256. Minimum Average Difference
# https://leetcode.com/problems/minimum-average-difference/

class Solution:
def minimumAverageDifference(self, nums: List[int]) -> int:
mmin = float('inf')
res = -1
n = len(nums)
curr = 0
s = sum(nums)

​
for i, x in enumerate(nums):
left = i + 1
right = n - i - 1

curr += x
s -= x
​
if i == n - 1:
ans = abs(floor(curr / left))
else:
ans = abs(floor(curr / left) - floor(s / right))
# print(i, ans)
if ans < mmin:
mmin = ans
res = i

return res



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).