SDE - 2

Design a system where user can upload, delete and share the video/image files. 
And can immediately view the uploaded files on the different logged-in system

You have to implement the pop function of Max Priority Queue and implement using a heap.

Functions :

a) push(int x) : 'x' has to be inserted in the priority queue. This has been implemented already

b) pop() : return the maximum element in the priority queue, if priority queue is empty then return '-1'.

Example:

We perform the following operations on an empty priority queue:

When operation push(5) is performed, we insert 1 in the priority queue.

When operation push(2) is performed, we insert 2 in the priority queue. 

When operation pop() is performed, we remove the maximum element from the priority queue and print which is 5.

When operation push(3) is performed, we insert 1 in the priority queue.

When operation pop() is performed, we remove the maximum element from the priority queue and print which is 3.

Ninja owns an electronic shop. In the shop, Ninja has 'N' bulbs. To sell these bulbs, Ninja has to check if they are of good quality. To check bulbs, Ninja uses a unique technique.

In this technique, in the first round he, turn on all the bulbs, then in the second round, he turns off every second bulb then in the third round, he chooses every third bulb and turns off it is on, or turn on if it is off and so no. He repeats this process 'N' times.

The number of bulbs that are on after the end of 'N' rounds are of good quality. Your task is to help Ninja in finding the number of good bulbs.

Example

Let 'N' = 4

Then in the first round, all bulbs are on.

In the second round bulbs, 2 and 4 are turned off.

In the third round bulb, 3 is turned off.

In the fourth round bulb, 4 is turned on.

So at the end of 'N' (4) round, only the bulb 1 and 4 is on, so there are 2 good bulbs.

You are given two sorted arrays of distinct integers, ‘ARR1’ and ‘ARR2’. If you find a common element in both arrays, you can switch from one array to another.

Your task is to find a path through the intersections i.e common integers of ‘ARR1’ and ‘ARR2’, that produces maximum sum and return that maximum sum as the answer.

For example:

Let ‘ARR1’ = [1, 5, 10, 15, 20]  and ‘ARR2’ = [2, 4, 5, 9, 15]
In this example, common points are 5, 15.

First, we start with ARR2 and take the sum till 5 (i.e. sum = 11). Then we will switch to ‘ARR1’ at element 10 and take the sum till 15. So sum = 36. Now no element is left in ‘ARR2’ after 15, so we will continue in array 1. Hence sum is 56. And the path is 2 -> 4 -> 5 -> 10 -> 15 -> 20.