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Therefore, the worst-case number of comparisons needed to select the second smallest is + ⌈ ⌉, the same number that would be obtained by holding a single-elimination tournament with a run-off tournament among the values that lost to the smallest value. However, the expected number of comparisons of a randomized selection algorithm can ...
The algorithm proceeds by finding the smallest (or largest, depending on sorting order) element in the unsorted sublist, exchanging (swapping) it with the leftmost unsorted element (putting it in sorted order), and moving the sublist boundaries one element to the right.
However, binary search can be used to solve a wider range of problems, such as finding the next-smallest or next-largest element in the array relative to the target even if it is absent from the array. There are numerous variations of binary search. In particular, fractional cascading speeds up binary searches for the same value in multiple arrays.
The first rightward pass will shift the largest element to its correct place at the end, and the following leftward pass will shift the smallest element to its correct place at the beginning. The second complete pass will shift the second largest and second smallest elements to their correct places, and so on.
Each of these integers can be stored with a word size of w, implying that . Data structures that solve the problem support these operations: [2] predecessor(x), which returns the largest element in S strictly smaller than x; successor(x), which returns the smallest element in S strictly greater than x
The simplest kind of query is to locate a record that has a specific field (the key) equal to a specified value v.Other common kinds of query are "find the item with smallest (or largest) key value", "find the item with largest key value not exceeding v", "find all items with key values between specified bounds v min and v max".
The second-smallest value y is T.children[T.aux.min].min, so it can be found in O(1) time. We delete y from the subtree that contains it. If x≠T.min and x≠T.max then we delete x from the subtree T.children[i] that contains x. If x == T.max then we will need to find the second-largest value y in the vEB tree and set T.max=y. We start by ...
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.