Calculate the number of slices. Slice means that the difference between the maximum and minimum in the slice <= 2.
eg. Given array 3 5 7 6 3.
There’re 10: (0,0) (1,1) (2,2) (3,3) (4,4) (5,5) (0,1) (1,2) (1,3) (2,3).
If only asked to output the total number of such slices, we can do it in O(n) using sliding-window-like algorithm, with the help of addtional queue-like data structure.
This is a very difficult question! Quoted from the top answer:
The basic idea is that you can start with arr and then see how many more elements you can include before you violate the max -min <= K constraint. When you reach some index j you can no longer include, you know the maximum subarray starting at index 0 is arr[0…j-1], and you can count j different subarrays there.
… you then start with a. Now [1…j-1] has to be valid, so you try to advance the right boundary again (try arr[1…j], then arr[1…j+1]) and so on until again you reach a point past which you can’t advance. Then you’ll try string at a, and so on. This is what they mean when they talk about “crawling” over the array.
The key issue here is how you will check whether a particular subarray is valid efficiently…
Sliding window is explained above. Now, the minMaxQueue explain in this pdf:
The PDF … have 1 queue each (i.e. 2 queues in total) for min and max (the min is treated analogous to the max case).
For maxQueue, whenever a new value comes in the max queue, they remove all values less than the value, since those can never be the max. For that reason the queue’s values are in descending order. And the max at any given time is the first element.
In this way, getting min/max is O(1), thus entire solution is O(n). Nice question!