Data Structures and Algorithm Analysis - Computer Science at ...

Sec. 5.5 Heaps and Priority Queues 175172 34645 6 7123 5(a)1723564 5 6 74 2 1 3(b)Figure 5.20 Two series of exchanges to build a max-heap. (a) This heap is builtby a series of nine exchanges in the order (4-2), (4-1), (2-1), (5-2), (5-4), (6-3),(6-5), (7-5), (7-6). (b) This heap is built by a series of four exchanges in the order(5-2), (7-3), (7-1), (6-1).Each call to insert takes Θ(log n) time in the worst case, because the valuebeing inserted can move at most the distance from the bottom of the tree to the topof the tree. Thus, to insert n values into the heap, if we insert them one at a time,will take Θ(n log n) time in the worst case.If all n values are available at the beginning of the building process, we canbuild the heap faster than just inserting the values into the heap one by one. ConsiderFigure 5.20(a), which shows one series of exchanges that could be used tobuild the heap. All exchanges are between a node and one of its children. The heapis formed as a result of this exchange process. The array for the right-hand tree ofFigure 5.20(a) would appear as follows:7 4 6 1 2 3 5Figure 5.20(b) shows an alternate series of exchanges that also forms a heap,but much more efficiently. The equivalent array representation would be7 5 6 4 2 1 3From this example, it is clear that the heap for any given set of numbers is notunique, and we see that some rearrangements of the input values require fewer exchangesthan others to build the heap. So, how do we pick the best rearrangement?