There is an O(N*logC) algorithm, where C is proportional to the maximum element value of the array. Comparing with some more complicated algorithms in recent papers, this algorithm is easier to understand, and can be implemented in a short time, and still fast enough in practical.
For simplicity, We assume there is at least one non-negative integers in the array.
The algorithm is based on binary search. At first, we can find that the final answer must be in the range [0, max(A)], and we half this interval in each iteration, until it is small enough (10-6 for example). In each iteration, assume the available interval is [a,b], we need to check whether the maximum average is no less than (a+b)/2. If so, we get a smaller interval [(a+b)/2, b], or else we get [a, (a+b)/2].
Now the problem is: Given a number K, how to check that the final answer is at least K?
Assume the average is at least K, there exist some i, j such that (A[i] + A[i+1] + ... + A[j]) / (j - i + 1) >= K. We multiply both sides by (j-i+1), and move the right side to left, and we get (A[i] - K) + (A[i+1] - K) + ... + (A[j] - K) >= 0.
So, let B[i] = A[i] - K, we only need to find an interval [i, j] (j - i + 1 > L) such that B[i] + ... + B[j] >= 0. Now the problem is: Given array B and length L, we are to find an interval of maximum sum whose length is more than L. If the maximum sum is >= 0, the original average number K is possible.
The second problem can be solved by linear scan. Let sumB[0] = 0, sumB[i] = B[1] + B[2] + ... + B[i]. For each index i, the max-sum interval which ended at B[i] is sumB[i] - min(sumB[0], sumB[1], ..., sumB[i-L-1]). When scanning the array with increasing i, we can maintain the min(sumB[0], ..., sumB[i-L-1]) on the fly.
The time complexity of the sub-problem is O(N). And we need O(logC) iterations, so the total complexity is O(N*logC).
P.s. This kinds of "average problem" belongs to a family of problems called fractional programming. The similar problems are minimum average-weighted spanning tree, minimum average-weighted cycle, etc.
P.s. again. The O(logC) is a loose bound. I think we can reduce it by some careful analysis.
