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 $O(1)$.

Python3JavaC++GoTypeScript

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

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;
    }
};

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;
}

Comments