Min Avg Two Slice

Problem

A non-empty array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a slice of array A (notice that the slice contains at least two elements). The average of a slice (P, Q) is the sum of A[P] + A[P + 1] + … + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + … + A[Q]) / (Q − P + 1).

For example, array A such that:

A[0] = 4
A[1] = 2
A[2] = 2
A[3] = 5
A[4] = 1
A[5] = 5
A[6] = 8

contains the following example slices:

• slice (1, 2), whose average is (2 + 2) / 2 = 2;
• slice (3, 4), whose average is (5 + 1) / 2 = 3;
• slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Write a function:

function solution(A);

that, given a non-empty array A consisting of N integers, returns the starting position of the slice with the minimal average. If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.

For example, given array A such that:

A[0] = 4
A[1] = 2
A[2] = 2
A[3] = 5
A[4] = 1
A[5] = 5
A[6] = 8

the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [2..100,000];
• each element of array A is an integer within the range [−10,000..10,000].

My Solution

failed

function solution(A) {
	let result = {};

	for (let i = 0; i < A.length - 2; i++) {
		for (let j = i + 2; j < A.length; j++) {
			let s = A.slice(i, j);
			let key = s.reduce((sum, cur) => sum + cur) / s.length;
			result[key] = i;
		}
	}

	return result[Math.min(...Object.keys(result))];
}

success O(N)

function solution(A) {
	var start = 0;

	var currentSum = A[0] + A[1];
	var minAvgSlice = currentSum / 2;
	for (var i = 2; i < A.length; i++) {
		currentSum += A[i];
		var newAvg = currentSum / 3;
		if (newAvg < minAvgSlice) {
			minAvgSlice = newAvg;
			start = i - 2;
		}

		currentSum -= A[i - 2];
		newAvg = currentSum / 2;
		if (newAvg < minAvgSlice) {
			minAvgSlice = newAvg;
			start = i - 1;
		}
	}

	return start;
}