Software Engineer

Given a DAG(direct acyclic graph), return the Topological Sorting of a given graph.

You have been given a long type array/list 'arr’ of size 'n’.

It represents an elevation map wherein 'arr[i]’ denotes the elevation of the 'ith' bar.

Print the total amount of rainwater that can be trapped in these elevations.

Note :

The width of each bar is the same and is equal to 1.

Example:

Input: ‘n’ = 6, ‘arr’ = [3, 0, 0, 2, 0, 4].

Output: 10

Explanation: Refer to the image for better comprehension:

Note :

You don't need to print anything. It has already been taken care of. Just implement the given function.

You are given a 'Nx3' 2-D array 'Jobs' describing 'N' jobs where 'Jobs[i][0]' denotes the id of 'i-th' job, 'Jobs[i][1]' denotes the deadline of 'i-th' job, and 'Jobs[i][2]' denotes the profit associated with 'i-th job'.

You will make a particular profit if you complete the job within the deadline associated with it. Each job takes 1 unit of time to be completed, and you can schedule only one job at a particular time.

Return the number of jobs to be done to get maximum profit.

Note :

If a particular job has a deadline 'x', it means that it needs to be completed at any time before 'x'.

Assume that the start time is 0.

For Example :

'N' = 3, Jobs = [[1, 1, 30], [2, 3, 40], [3, 2, 10]].

All the jobs have different deadlines. So we can complete all the jobs.

At time 0-1, Job 1 will complete.
At time 1-2, Job 3 will complete.
At time 2-3, Job 2 will complete.

So our answer is [3 80].

Given an array/list of length ‘n’, where the array/list represents the boards and each element of the given array/list represents the length of each board. Some ‘k’ numbers of painters are available to paint these boards. Consider that each unit of a board takes 1 unit of time to paint.

You are supposed to return the area of the minimum time to get this job done of painting all the ‘n’ boards under a constraint that any painter will only paint the continuous sections of boards.

Example :

Input: arr = [2, 1, 5, 6, 2, 3], k = 2

Output: 11

Explanation:
First painter can paint boards 1 to 3 in 8 units of time and the second painter can paint boards 4-6 in 11 units of time. Thus both painters will paint all the boards in max(8,11) = 11 units of time. It can be shown that all the boards can't be painted in less than 11 units of time.

There are ‘N’ houses in a village. Ninja wants to supply water for all the houses by building wells and laying pipes.

For each house ‘i’, we can either build a well inside it directly with cost ‘WELLS[i]’, or pipe in water from another well to it. The total cost to lay pipes between houses is given by the array ‘PIPES’, where ‘PIPES[i]’ = ‘[HOUSE1, HOUSE2, COST]’ and the ‘COST’ represent the total cost connect ‘HOUSE1’ and ‘HOUSE2’ together using a pipe.

Note: Given all the connections are bidirectional.

For Example:

For ‘N’ = 3, ‘WELLS[]’ = ‘[1,2,2]’, ‘PIPES[]’ = [ [1, 2, 1], [2 , 3, 1]]. The image shows the costs of connecting houses using pipes. The best strategy is to build a well in the first house with cost 1 and connect the other houses to it with cost 2 so the total cost is 3.

Ninja wants to find out the minimum total cost to supply water to all houses in the village. Can you help the Ninja to find out this minimum cost?

Given an array of integers ‘ARR’ and an integer ‘X’, you are supposed to find the number of subarrays of 'ARR' which have bitwise XOR of the elements equal to 'X'.

Note:

An array ‘B’ is a subarray of an array ‘A’ if ‘B’ that can be obtained by deletion of, several elements(possibly none) from the start of ‘A’ and several elements(possibly none) from the end of ‘A’.