HackerRank List Comprehensions solution in Python
2 min read

HackerRank List Comprehensions solution in Python

HackerRank List Comprehensions solution in Python

Let's learn about list comprehensions! You are given three integers  X, Y and Z representing the dimensions of a cuboid along with an integer N. You have to print a list of all possible coordinates given by  (i,j,k) on a 3D grid where the sum of  (i+j+k) is not equal to N.

Input Format

Four integers X, Y, Z  and N each on four separate lines, respectively.

Constraints

Print the list in lexicographic increasing order.

Sample Input 0

1

1

1

2

Sample Output 0

[[0, 0, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1]]

Explanation 0

Concept

You have already used lists in previous hacks. List comprehensions are an elegant way to build a list without having to use different for loops to append values one by one. This example might help.

Example: You are given two integers x and y . You need to find out the ordered pairs ( i , j ) , such that ( i + j ) is not equal to n and print them in lexicographic order.( 0 <= i <= x ) and ( 0 <= j <= y) This is the code if we dont use list comprehensions in Python.

python x = int ( raw_input()) y = int ( raw_input()) n = int ( raw_input()) ar = [] p = 0 for i in range ( x + 1 ) : for j in range( y + 1): if i+j != n: ar.append([]) ar[p] = [ i , j ] p+=1 print ar
Other smaller codes may also exist, but using list comprehensions is always a good option. Code using list comprehensions:

python x = int ( raw_input()) y = int ( raw_input()) n = int ( raw_input()) print [ [ i, j] for i in range( x + 1) for j in range( y + 1) if ( ( i + j ) != n )]

Sample Input 1

2

2

2

2

Sample Output 1[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 2], [0, 2, 1], [0, 2, 2], [1, 0, 0], [1, 0, 2], [1, 1, 1], [1, 1, 2], [1, 2, 0], [1, 2, 1], [1, 2, 2], [2, 0, 1], [2, 0, 2], [2, 1, 0], [2, 1, 1], [2, 1, 2], [2, 2, 0], [2, 2, 1], [2, 2, 2]]

Solution in Python3

x,y,z,n = [int(input()) for i in range(4)]
print([[i,j,k] for i in range(x+1) for j in range(y+1) for k in range(z+1) if ((i+j+k) != n)])

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