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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.
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 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.
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.
Related problems include approximate sorting (sorting a sequence to within a certain amount of the correct order), partial sorting (sorting only the k smallest elements of a list, or finding the k smallest elements, but unordered) and selection (computing the kth smallest element). These can be solved inefficiently by a total sort, but more ...
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 ...
The best you can do is (in case of array implementation) simply concatenating the two heap arrays and build a heap of the result. [13] A heap on n elements can be merged with a heap on k elements using O(log n log k) key comparisons, or, in case of a pointer-based implementation, in O(log n log k) time. [14]
Firstly, computing median of an odd list is faster and simpler; while one could use an even list, this requires taking the average of the two middle elements, which is slower than simply selecting the single exact middle element. Secondly, five is the smallest odd number such that median of medians works.