LeetCode 2256. Minimum Average Difference Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)