# Count spread in O(n)

This is not hard but still interesting.

A spread of an array of numbers is the difference between its maximum and minimum. Given an array of numbers \(x_1,\ldots,x_n\), compute the sum of spreads of all of its continuous subarrays in \(O(n)\) time.

Solution:

If we could compute the sum of maximum of all continuous subarrays then it's easy. To do this, suppose we know this sum for \(x_1,\ldots,x_{n-1}\). When \(x_n\) comes in, we want to add to this sum the sum of maximums of \((x_1,\ldots,x_n),(x_2,\ldots,x_n),\ldots,(x_{n-1},x_n),(x_n)\), or \(y_1+y_2+\ldots+y_n\), where \(y_k\) is the maximum between \(x_k\) and the last number. Clearly \(y\) never goes up, so we could use a stack of triple \((value, right, area)\) to store info we need. \((value, right, area)\) is such that for \(right_{k-1} + 1\leq i\leq right_k\), \(y_i=value_k\), and the cumulative area of \(y\) curve from the beginning to \(right_k\) is \(area_k\).

After \(x_n\) comes in, \((value, right, area)\) is updated as this. Pop out all entries where \(value\lt x_n\). Now if the top of the stack has \(value=x_n\), update its \(area\) and \(right\) accordingly. Otherwise push a new entry with \(value=x_n\) and \(right=n\). Its \(area\) should be the preceding \(area\) plus \(x_n\times\delta\) where \(\delta\) is \(n\) minus the preceding \(right\).

After \((value, right, area)\) is updated, add the last \(area\) to the sum. The \(pop\) takes amortized \(O(1)\) per element, and the rest takes a fixed \(O(1)\).

post by Shen-Fu Tsai