2256. Minimum Average Difference

Description

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 = [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

Solutions

Solution 1: Traverse

We directly traverse the array $nums$. For each index $i$, we maintain the sum of the first $i+1$ elements $pre$ and the sum of the last $n-i-1$ elements $suf$. We calculate the absolute difference of the average of the first $i+1$ elements and the average of the last $n-i-1$ elements, denoted as $t$. If $t$ is less than the current minimum value $mi$, we update the answer $ans=i$ and the minimum value $mi=t$.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

• class Solution {
public int minimumAverageDifference(int[] nums) {
int n = nums.length;
long pre = 0, suf = 0;
for (int x : nums) {
suf += x;
}
int ans = 0;
long mi = Long.MAX_VALUE;
for (int i = 0; i < n; ++i) {
pre += nums[i];
suf -= nums[i];
long a = pre / (i + 1);
long b = n - i - 1 == 0 ? 0 : suf / (n - i - 1);
long t = Math.abs(a - b);
if (t < mi) {
ans = i;
mi = t;
}
}
return ans;
}
}

• class Solution {
public:
int minimumAverageDifference(vector<int>& nums) {
int n = nums.size();
using ll = long long;
ll pre = 0;
ll suf = accumulate(nums.begin(), nums.end(), 0LL);
int ans = 0;
ll mi = suf;
for (int i = 0; i < n; ++i) {
pre += nums[i];
suf -= nums[i];
ll a = pre / (i + 1);
ll b = n - i - 1 == 0 ? 0 : suf / (n - i - 1);
ll t = abs(a - b);
if (t < mi) {
ans = i;
mi = t;
}
}
return ans;
}
};

• class Solution:
def minimumAverageDifference(self, nums: List[int]) -> int:
pre, suf = 0, sum(nums)
n = len(nums)
ans, mi = 0, inf
for i, x in enumerate(nums):
pre += x
suf -= x
a = pre // (i + 1)
b = 0 if n - i - 1 == 0 else suf // (n - i - 1)
if (t := abs(a - b)) < mi:
ans = i
mi = t
return ans


• func minimumAverageDifference(nums []int) (ans int) {
n := len(nums)
pre, suf := 0, 0
for _, x := range nums {
suf += x
}
mi := suf
for i, x := range nums {
pre += x
suf -= x
a := pre / (i + 1)
b := 0
if n-i-1 != 0 {
b = suf / (n - i - 1)
}
if t := abs(a - b); t < mi {
ans = i
mi = t
}
}
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function minimumAverageDifference(nums: number[]): number {
const n = nums.length;
let pre = 0;
let suf = nums.reduce((a, b) => a + b);
let ans = 0;
let mi = suf;
for (let i = 0; i < n; ++i) {
pre += nums[i];
suf -= nums[i];
const a = Math.floor(pre / (i + 1));
const b = n - i - 1 === 0 ? 0 : Math.floor(suf / (n - i - 1));
const t = Math.abs(a - b);
if (t < mi) {
ans = i;
mi = t;
}
}
return ans;
}