Given an array, find subarray that has one max possible value defined by subarray length times minimum

Question

Let A be an array with N entries. We consider continuous subarrays of it, these must be continuous , meaning they start at some index I and end on some index J. Denote m(I,J) the minimum element of the subarray starting at I and ending on J (contatining elements at index I and J inclusive). Find the maximum among m(I,J)*(J-I+1). J-I+1 is the subarray length.

Instead of running a loop over I and J and look for which has the minimum there should be a better algorithm in O(N) time. How can I do it? I am thinking about a thing like sliding window... First I pick the first element, I sucessively increase the width of the sliding window as long as the first element remains subarray minimum. That means that I will only have greater values. And I would eventually try to get rid of first element from sliding window to check whether greater elements lead to greater results. But how do I do it precisely?

Answer 1

1. Traverse the array. For each element, calculate the index of the next smaller element on both sides, and store the result in arrays left[] and right[].
2. Let the element at index i be m. For any subarray [l, r] where left[i] < l <= i and i <= r < right[i], m will be it's smallest element. Thus, of all subarrays having array[i] as the smallest element, the subarray of maximum length would be the one between (left[i], right[i]). So it's value would be array[i] * (right[i] - left[i] - 1).
3. Calculate the global maximum of this value over the array. That would be your answer.

Edit:

This logic works for all i where array[i] >= 0. If array[i] < 0, we'll pick the shortest subarray possible for it, which will have length 1. So the value produced for it will be array[i].