Assistant System Engineer

Given a matrix ‘UNIVERSE’ with 3 rows and ‘N’ columns, with the characters { # , * , . } and these characters represent a cluster of stars and galaxies in space. Stars are represented by ‘*’ symbol. The ‘.’ symbol represents empty space.

We define a constellation as a 3x3 matrix which contains stars in the shape of vowels. A group of constellations is defined as a galaxy. Note that a galaxy may contain many stars but they will never be overlapping. Two galaxies are separated by a column of ‘#’.

Given the ‘UNIVERSE’ matrix, print a string which denotes all the vowels formed by stars and ‘#’ present in the matrix.

A looks like this :

. * .
* * *
* . *

E looks like this :

* * *
* * *
* * *

I looks like this :

* * *
. * .
* * *

O looks like this :

* * *
* . *
* * *

U looks like this :

* . *
* . *
* * *

Note :

It is guaranteed that no two constellations are overlapping.

Ninja is in space with his super spaceship having unlimited fuel. Ninja initially has a health level ‘H’ and his spaceship has an armour ‘A’. He decides to play a game, where at any instant he can be only on one of the 3 planets: VENUS, MARS or SATURN. At every unit of time, he has to change his location and you can assume he can change his location instantly. Each planet has different effects on health and armour which are :

Venus: Decreases health by 4, decreases armour by 8.
Mars: Increases health by 3, increases armour by 2. 
Saturn: Decreases health by 10, increases armour by 5. 

Now the game will end if either ‘H’ <= 0 or ‘A’ <= 0.

Your task is to find the maximum time Ninja can survive.

Note :

You can choose any of the 3 planets during your first move.

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.

You are given an array 'ARR' of 'N' integers and two integers 'K' and 'M'.

You need to return true if the given array can be divided into pairs such that the sum of every pair gives remainder 'M' when divided by 'K'. Otherwise, you need to return false.

For example:

If the given array is [2, 1, 5, 7] and K = 9 and M = 3. Then you need to return true because we can divide the array into two pairs, i.e (2, 1) and (5, 7) whose sums are 3 and 12, which when divided by 9 gives remainder 3, thus it is possible to divide the given array into pairs.  

Note:

Every element of the array should contribute to only one pair, i.e if the array is [3, 0, 0] and K = 2 and M = 1, then you need to return false, as element 3 will make a pair with any one of the 0. 

You have been given two integers ‘DAY_HOURS’ and ‘PARTS’. Where ‘DAY_HOURS’ is the number of hours in a day and a day can be divided into ‘PARTS’ equal parts. Your task is to find total instances of equivalent prime groups. Prime group refers to the group of elements (hours) which are prime and measure the same equivalent time since the start of the day.

For example, if we consider ‘DAY_HOURS’ to be 24 and ‘PARTS’ to be 2, then the day of total 24 hours is divided into 2 parts ( 1 - 12 ) and ( 13 - 24 ). 5 hours in the first part of the day is equivalent to 17, which is 5 hours into the second part of the day. And since 5 and 17 both are prime, they can be considered as a prime group.

Note:

1. Day starts with hour 1 and ends with hour  ‘DAY_HOURS’.

2. Each hour of the prime group should be in a different part of the day.

3. If there is no prime group then return zero.

4. ‘DAY_HOURS’ should be divisible by ‘PARTS’, meaning that the number of hours per part (DAY_HOURS/PARTS)  should be a natural number.

Example:

Let ‘DAY_HOURS’ = 20  and ‘PARTS’ = 2

Hence the view of our day would be in the following format: 

1  2  3  4  5  6  7  8  9 10      -  Part 1
11 12 13 14 15 16 17 18 19 20     -  Part 2

 1-11  Not a prime group because 1 is not prime.
 2-12  Not a prime group because 12 is not prime.
 3-13  Because both 3 and 13 are prime, it is an equivalent prime group.
 4-14  Not a prime group because 4 and 14 are not prime.
 5-15  Not a prime group because 15 is not prime.
 6-16  Not a prime group, because 6 and 16 are not prime.
 7-17  Because both 7 and 17 are prime, it is an equivalent prime group.
 8-18  Not a prime group, because 8 and 18 are, is not prime.
 9-19  Not a prime group because 9 is not prime.
 10-20 Not a prime group because both 10 and 20 are not prime.

 Hence there are 2 equivalent prime groups in the above format which are 3-13 and 7-17.

You have been given two integers ‘DAY_HOURS’ and ‘PARTS’. Where ‘DAY_HOURS’ is the number of hours in a day and a day can be divided into ‘PARTS’ equal parts. Your task is to find total instances of equivalent prime groups. Prime group refers to the group of elements (hours) which are prime and measure the same equivalent time since the start of the day.

For example, if we consider ‘DAY_HOURS’ to be 24 and ‘PARTS’ to be 2, then the day of total 24 hours is divided into 2 parts ( 1 - 12 ) and ( 13 - 24 ). 5 hours in the first part of the day is equivalent to 17, which is 5 hours into the second part of the day. And since 5 and 17 both are prime, they can be considered as a prime group.

Note:

1. Day starts with hour 1 and ends with hour  ‘DAY_HOURS’.

2. Each hour of the prime group should be in a different part of the day.

3. If there is no prime group then return zero.

4. ‘DAY_HOURS’ should be divisible by ‘PARTS’, meaning that the number of hours per part (DAY_HOURS/PARTS)  should be a natural number.

Example:

Let ‘DAY_HOURS’ = 20  and ‘PARTS’ = 2

Hence the view of our day would be in the following format: 

1  2  3  4  5  6  7  8  9 10      -  Part 1
11 12 13 14 15 16 17 18 19 20     -  Part 2

 1-11  Not a prime group because 1 is not prime.
 2-12  Not a prime group because 12 is not prime.
 3-13  Because both 3 and 13 are prime, it is an equivalent prime group.
 4-14  Not a prime group because 4 and 14 are not prime.
 5-15  Not a prime group because 15 is not prime.
 6-16  Not a prime group, because 6 and 16 are not prime.
 7-17  Because both 7 and 17 are prime, it is an equivalent prime group.
 8-18  Not a prime group, because 8 and 18 are, is not prime.
 9-19  Not a prime group because 9 is not prime.
 10-20 Not a prime group because both 10 and 20 are not prime.

 Hence there are 2 equivalent prime groups in the above format which are 3-13 and 7-17.