Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 10, Problem 2P
(a)
Program Plan Intro
To implement mergeable heap using lined list for sorted lists and analyze the running time complexity for operations: INSERT, MINIMUM, EXTRACT-MIN and UNION.
(b)
Program Plan Intro
To implement mergeable heap using lined list for unsorted lists and analyze the running time complexity for operations: INSERT, MINIMUM, EXTRACT-MIN and UNION.
(c)
Program Plan Intro
To implement mergeable heap using lined list for unsorted lists and disjoint dynamic sets and analyze the running time complexity for operations: INSERT, MINIMUM, EXTRACT-MIN and UNION.
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void insert(int p);
// Insert adds the element p to the heap.
//
// In this heap the element is an int and IS the priority.
//
// We do not know at this abstract level if this will be
// a min-heap or a max-heap (or even something else like
// a fibonacci heap) and those details will be added in
// the class that implements this heap interface.
// Duplicate values are allowed.
//
// Error: The insert should fail if the heap is "full"
// meaning the array is maxed out and there is no room
// in the array for another element. In this case
// simply return, and do nothing.
//Min-heap example:
// Suppose we have the following min-heap:
// 3, 5, 4, 12, 6, 9
// After inserting 2, the min-heap now looks as follows:
// 2, 5, 3, 12, 6, 9, 4
Your task is to implement the methods below in MinHeap.java for a min-heap using the instructions in Heap.java. You should not modify the code in Heap.java.
Hint:…
Object is name(age,time)
For example : apple (1,11)
the Binary tree order depend on age
Write a java code that would add a new node to the tree in the leaf position where binary search determines a node for d should be inserted.
To build fix the trees so that the properties of the max heap are maintained (i.e. the parent node must have higher adoption priority than its children). This means that we need to perform upheap if needed. Note that since this is also a binary search tree, we need to make sure that when performing upheap we don’t break the properties of the binary search tree. To ensure this, instead of performing upheap as seen in class, we will need to implement a tree rotation that reverses the parent-child relationship whenever necessary. Depending if the child that has to be swap in the parent position is the left or the right child, we will need to perform a right rotation or a left rotation
6.
Idea: Maintain a max heap of k elements.We can iterate through all points.If a point p has a smaller distance to the origin than the top element of aheap, we add point p to the heap and remove the top element.After iterating through all points, our heap contains the k closest points tothe origin."""
from heapq import heapify, heappushpop
def k_closest(points, k, origin=(0, 0)): # Time: O(k+(n-k)logk) # Space: O(k) """Initialize max heap with first k points. Python does not support a max heap; thus we can use the default min heap where the keys (distance) are negated. """ heap = [(-distance(p, origin), p) for p in points[:k]] heapify(heap)
""" For every point p in points[k:], check if p is smaller than the root of the max heap; if it is, add p to heap and remove root. Reheapify. """ for point in points[k:]: dist = distance(point, origin)
heappushpop(heap, (-dist, point)) # heappushpop does conditional check """Same as:…
Chapter 10 Solutions
Introduction to Algorithms
Ch. 10.1 - Prob. 1ECh. 10.1 - Prob. 2ECh. 10.1 - Prob. 3ECh. 10.1 - Prob. 4ECh. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.2 - Prob. 1ECh. 10.2 - Prob. 2ECh. 10.2 - Prob. 3E
Ch. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - Prob. 7ECh. 10.2 - Prob. 8ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.4 - Prob. 1ECh. 10.4 - Prob. 2ECh. 10.4 - Prob. 3ECh. 10.4 - Prob. 4ECh. 10.4 - Prob. 5ECh. 10.4 - Prob. 6ECh. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3P
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