Foundations of Data Structures Practical Session #8 Heaps.

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Presentation transcript:

Foundations of Data Structures Practical Session #8 Heaps

Binary heap properties Heap A binary heap can be considered as a complete binary tree (the last level is full from the left to a certain point). Maximum- Heap greater maximal Each node is greater than or equal to each of its children. The root in a Maximum-Heap is the maximal element in the heap. Minimum- Heap greater minimal Each node is greater than or equal to each of its children. The root in a Minimum-Heap is the minimal element in the heap. Operations (On max- heap) 2

Heap-array

Insert (heapify-up) 1.Add the element to the bottom level of the heap. 2.Compare the added element with its parent; if they are in the correct order, stop. 3.If not, swap the element with its parent and return to the previous step Insert 152. Swap 15, 83. Swap 15, 11

Delete root (heapify-down) 1.Replace the root of the heap with the rightmost element on the last level. 2.Compare the new root with its children; if they are in the correct order, stop. 3.If not, swap the element with one of its children and return to the previous step. (Swap with its smaller child in a min- heap and its larger child in a max-heap.) 5 1. Remove 112. Put 4 as root3. Swap 4, 8

Question 1 A max-heap is given as a heap-array A of size 63. The heap contains all keys in the range , each key exactly ones. 1.Where can be located the key 72? Specify possible indexes. Answer: A[1]. in a max-heap the maximum is always at the root. 2.Where can be located the key 10? Answer: anywhere on the last level, i.e., in A[32]…A[63]. 3.Which keys can be located in A[2]? Answer: The key 40 can be located at A[2] when all the larger keys, are rooted at key 71, located at A[3]. 4.Where can be located the key 70? Answer: in A[2]…A[7]. There are exactly two elements larger than 70 (71, 72), so it can’t reside lower than level 2. On level two it can reside anywhere. 6

Question 2 1.Given a maximum-heap H, what is the time complexity of finding the 3 largest keys in H? 2.In general, what is the time complexity of finding the C largest keys in H? (Where C is a parameter) 7

Question 2 solution It takes O(1) to find the 3 maximal keys in H. The 3 maximal keys in H are: The root. The root's maximal son, y. The largest among y's maximal son and root's other son x. 8

Question 2 solution 9

Question 3 10InitInsert(x) FindMin() FindMax DelMin DelMax

Question 3 solution Store the elements in two heaps, Min-Heap H min and Max-Heap H max with mutual pointers, each element has a pointer to the element with the same value in the other heap. 11 Init Build H min in O(n) and H max in O(n) Insert(x) Insert x to H min in O(logn) Insert x to H max in O(logn) Update the pointers in O(1) FindMin() Return the root of H min in O(1) FindMax Return the root of H max in O(1) DelMin Delete the minimum from H min in O(logn) Delete the same element from H max by following the mutual pointer in O(logn) DelMax Delete the maximum from H max in O(logn) Delete the same element from H min by following the mutual pointer in O(logn)

Question 4 Two min-heaps are given, H 1 and H 2 with n 1 and n 2 keys respectively. Each element of H 1 is smaller than each element of H 2. Suggest an algorithm for merging H 1 and H 2 into a single heap H (with n 1 +n 2 keys) in O(n 2 ) time. 12

Question 4 solution 13

Question 4 solution 14

Question 4 solution 15

Question 5 16 Delete(H, i) if (heap_size(H) < i) error(“heap underflow”) key = H[i] H[i]= H[heap_size] remove H[heap_size] heap_size-- Down-Heapify(H, i) Up-Heapify(H, i)