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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 = [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 (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;
    }
}

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

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

  • 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


  • typedef long long ll;

class Solution {
public:
    int minimumAverageDifference(vector<int>& nums) {
        int n = nums.size();
        vector<ll> s(n);
        s[0] = nums[0];
        for (int i = 1; i < n; ++i) s[i] = s[i - 1] + nums[i];
        int ans = 0;
        ll mi = LONG_MAX;
        for (int i = 0; i < n; ++i) {
            ll a = s[i] / (i + 1);
            ll b = i == n - 1 ? 0 : (s[n - 1] - s[i]) / (n - i - 1);
            ll t = abs(a - b);
            if (mi > t) {
                ans = i;
                mi = t;
            }
        }
        return ans;
    }
};

  • func minimumAverageDifference(nums []int) int {
	n := len(nums)
	s := make([]int, n)
	s[0] = nums[0]
	for i := 1; i < n; i++ {
		s[i] = s[i-1] + nums[i]
	}
	ans := 0
	mi := math.MaxInt32
	for i := 0; i < n; i++ {
		a := s[i] / (i + 1)
		b := 0
		if i != n-1 {
			b = (s[n-1] - s[i]) / (n - i - 1)
		}
		t := abs(a - b)
		if mi > t {
			ans = i
			mi = t
		}
	}
	return ans
}

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;
}

Explain:

nope.

Complexity:

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

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