A CodeFair Challenge

Solution to the Ashesi Code Fair 2015 Competition (Advance Problem) in Php

I took 1st runner up and I was suprised when the Judges said my code ran for 28secs Guys, help me confirm that

TL;DR

View Code on github

Here is the Problem:

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:

int solution(int A[], int N);

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.