A non-empty zero-indexed 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:

class Solution { public int solution(int[] A); }

that, given a non-empty zero-indexed 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.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

This is my best solution, but obviously not optimal in terms of time complexity.

Any ideas?

public int solution(int[] A) { int result = 0; int N = A.length; int [] prefix = new int [N+1]; for (int i = 1; i < prefix.length; i++) { prefix[i] = prefix[i-1] + A[i-1]; } double avg = Double.MAX_VALUE; for (int i = 1; i < N; i++) { for (int j = i+1; j <=N; j++) { double temp = (double)(prefix[j]-prefix[i-1]) /(double)(j-i+1); if (temp < avg) { avg = temp; result = i-1; } } } return result; }

https://codility.com/demo/results/demo65RNV5-T36/

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## Answer

I had posted this some days ago:

Check this out:

http://codesays.com/2014/solution-to-min-avg-two-slice-by-codility/

In there, they explain with great detail why their solution works. I haven’t implemented it myself yet, but I will definitely try it.

Hope it helps!

but I just saw it was deleted by a moderator. They say the link is dead, but I just tried it and it works fine. I’m posting it once again, hoping it can be validated that the link is good.

And now I can also provide my implementation, based on the codesays link that I provided before: https://codility.com/demo/results/demoERJ4NR-ETT/

class Solution { public int solution(int[] A) { int minAvgIdx=0; double minAvgVal=(A[0]+A[1])/2; //At least two elements in A. double currAvg; for(int i=0; i<A.length-2; i++){ /** * We check first the two-element slice */ currAvg = ((double)(A[i] + A[i+1]))/2; if(currAvg < minAvgVal){ minAvgVal = currAvg; minAvgIdx = i; } /** * We check the three-element slice */ currAvg = ((double)(A[i] + A[i+1] + A[i+2]))/3; if(currAvg < minAvgVal){ minAvgVal = currAvg; minAvgIdx = i; } } /** * Now we have to check the remaining two elements of the array * Inside the for we checked ALL the three-element slices (the last one * began at A.length-3) and all but one two-element slice (the missing * one begins at A.length-2). */ currAvg = ((double)(A[A.length-2] + A[A.length-1]))/2; if(currAvg < minAvgVal){ minAvgVal = currAvg; minAvgIdx = A.length-2; } return minAvgIdx; } }