List comprehension python Hackerrank

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. 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. Here, 0 <= i <= x; 0 <= j <= y; 0 <= k <= z. Please use list comprehensions rather than multiple loops, as a learning exercise.

Example

x = 1
y = 1
z = 2
n = 3

All permutations of [i, j, k] are:

[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 1, 0], [0, 1, 1], [0, 1, 2], [1, 0, 0], [1, 0, 1], [1, 0, 2], [1, 1, 0], [1, 1, 1], [1, 1, 2]].

Print an array of the elements that do not sum to n = 3.

[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1],  [1, 1, 0],  [1, 1, 2]]

Input Format

Four integers x, y, z and n, each on a separate line.

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

Each variable x, y and z will have values of 0 or 1. All permutations of lists in the form [i, j, k] = [[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1],  [1, 1, 0]].

Remove all arrays that sum to n = 2 to leave only the valid permutations.

List Comprehensions in Python – Hacker Rank Solution

if __name__ == '__main__':

x = int(input())

y = int(input())

z = int(input())

n = int(input())

output = [];

abc = [];

for X in range(x+1):

for Y in range(y+1):

for Z in range(z+1):

if X+Y+Z != n:

abc = [X,Y,Z];

output.append(abc);

print(output);