Define max and min heap, also the search time of heap.
Answer / Megha Agarwal
"Max Heap and Min Heap are types of binary heaps used in data structures.n1. Max Heap: A Max Heap is a complete binary tree with the property that the parent node is greater than or equal to its child nodes. The maximum value is always at the root.n2. Min Heap: A Min Heap is a complete binary tree with the property that the parent node is less than or equal to its child nodes. The minimum value is always at the root.n3. Search time for both max and min heaps in the average case is O(log n)."
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3.2 Consider the following class: public class Point { protected int x, y; public Point(int xx, int yy) { x = xx; y = yy; } public Point() { this(0, 0); } public int getx() { return x; } public int gety() { return y; } public String toString() { return "("+x+", "+y+")"; } } Say you wanted to define a rectangle class that stored its top left corner and its height and width as fields. 3.2.1 Why would it be wrong to make Rectangle inherit from Point (where in fact it would inherit the x and y coordinates for its top left corner and you could just add the height and width as additional fields)? (1) 8 Now consider the following skeleton of the Rectangle class: public class Rectangle { private Point topLeft; private int height, width; public Rectangle(Point tl, int h, int w) { topLeft = tl; height = h; width = w; } public Rectangle() { this(new Point(), 0, 0); } // methods come here } 3.2.2 Explain the no-argument constructor of the Rectangle class given above. 3.2.3 Write methods for the Rectangle class to do the following: • a toString() method that returns a string of the format "top left = (x, y); height = h; width = w " where x, y, h and w are the appropriate integer values. • an above() method that tests whether one rectangle is completely above another (i.e. all y values of the one rectangle are greater than all y values of the other). For example, with the following declarations Rectangle r1 = new Rectangle(); Rectangle r2 = new Rectangle(new Point(2,2), 1, 4); the expression r2.above(r1) should give true, and r2.above (r2) should give false. (You can assume that the height of a rectangle is never negative.) (2) (5)
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