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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 …
The best case input is an array that is already sorted. In this case insertion sort has a linear running time (i.e., O(n)). During each iteration, the first remaining element of the input is only compared with the right-most element of the sorted subsection of the array. The simplest worst case input is an array sorted in reverse order.
Given a function that accepts an array, a range query (,) on an array = [,..,] takes two indices and and returns the result of when applied to the subarray [, …,].For example, for a function that returns the sum of all values in an array, the range query (,) returns the sum of all values in the range [,].
A further relaxation requiring only a list of the k smallest elements, but without requiring that these be ordered, makes the problem equivalent to partition-based selection; the original partial sorting problem can be solved by such a selection algorithm to obtain an array where the first k elements are the k smallest, and sorting these, at a total cost of O(n + k log k) operations.
The bin packing problem can also be seen as a special case of the cutting stock problem. When the number of bins is restricted to 1 and each item is characterized by both a volume and a value, the problem of maximizing the value of items that can fit in the bin is known as the knapsack problem .
Our second gap (k) is 256, which breaks the array into four sections (starting at 0, 256, 512, 768), and we make sure the first items in each section are sorted relative to each other, then the second item in each section, and so on. In practice the gap sequence could be anything, but the last gap is always 1 to finish the sort (effectively ...
The problem is then to produce an array such that all "bottom" elements come before (have an index less than the index of) all "middle" elements, which come before all "top" elements. One algorithm is to have the top group grow down from the top of the array, the bottom group grow up from the bottom, and keep the middle group just above the ...
Here input is the input array to be sorted, key returns the numeric key of each item in the input array, count is an auxiliary array used first to store the numbers of items with each key, and then (after the second loop) to store the positions where items with each key should be placed, k is the maximum value of the non-negative key values and ...