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Range minimum query reduced to the lowest common ancestor problem. Given an array A[1 … n] of n objects taken from a totally ordered set, such as integers, the range minimum query RMQ A (l,r) =arg min A[k] (with 1 ≤ l ≤ k ≤ r ≤ n) returns the position of the minimal element in the specified sub-array A[l … r].
As a baseline algorithm, selection of the th smallest value in a collection of values can be performed by the following two steps: . Sort the collection; If the output of the sorting algorithm is an array, retrieve its th element; otherwise, scan the sorted sequence to find the th element.
Selecting the minimum requires scanning elements (taking comparisons) and then swapping it into the first position. Finding the next lowest element requires scanning the remaining elements (taking comparisons) and so on. Therefore, the total number of comparisons is
In computer science, the range query problem consists of efficiently answering several queries regarding a given interval of elements within an array. For example, a common task, known as range minimum query , is finding the smallest value inside a given range within a list of numbers.
[37] [38] Exchange sort works by comparing the first element with all elements above it, swapping where needed, thereby guaranteeing that the first element is correct for the final sort order; it then proceeds to do the same for the second element, and so on. It lacks the advantage that bubble sort has of detecting in one pass if the list is ...
Extracting the minimum-priority element, in the basic version of the bucket queue, searches from the start of to find its first non-empty element: [] is empty but [] = {}, a non-empty set. It chooses an arbitrary element of this set (the only element, y {\displaystyle y} ) as the minimum-priority element.
The procedure for deleting the root from the heap (effectively extracting the maximum element in a max-heap or the minimum element in a min-heap) while retaining the heap property is as follows: Replace the root of the heap with the last element on the last level. Compare the new root with its children; if they are in the correct order, stop.
One of the two elements in the second level, which is a max (or odd) level, is the greatest element in the min-max heap Let x {\displaystyle x} be any node in a min-max heap. If x {\displaystyle x} is on a min (or even) level, then x . k e y {\displaystyle x.key} is the minimum key among all keys in the subtree with root x {\displaystyle x} .