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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 …
For example, for the array of values [−2, 1, −3, 4, −1, 2, 1, −5, 4], the contiguous subarray with the largest sum is [4, −1, 2, 1], with sum 6. Some properties of this problem are: If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array.
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 [,].
The k-way merge problem consists of merging k sorted arrays to produce a single sorted array with the same elements. Denote by n the total number of elements. n is equal to the size of the output array and the sum of the sizes of the k input arrays. For simplicity, we assume that none of the input arrays is empty.
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 ...
When the array contains only duplicates of a relatively small number of items, a constant-time perfect hash function can greatly speed up finding where to put an item 1, turning the sort from Θ(n 2) time to Θ(n + k) time, where k is the total number of hashes. The array ends up sorted in the order of the hashes, so choosing a hash function ...
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The array of Fibonacci numbers is defined where F k+2 = F k+1 + F k, when k ≥ 0, F 1 = 1, and F 0 = 1. To test whether an item is in the list of ordered numbers, follow these steps: Set k = m. If k = 0, stop. There is no match; the item is not in the array. Compare the item against element in F k−1. If the item matches, stop.