FAANG Interview Question : Arrangement of Bricks

Hello reader, this is my first post on Medium. I hope you enjoy it!

I was asked the below question in one of the coding interviews, here is my approach and solution for the same.

Question:

We are given a wall of height h and length l and we have to design it with bricks in such a way that:

1. When the bricks are stacked on top of each other no two bricks’ edges coincide(except at the corners).
2. After stacking the bricks the total length is l and height is h.
3. There are 2 kinds of bricks: 2X1 and 3X1.

Example:

Wall 1 = length:7 and height:5

Wall 2 = length:6 and height:4

Observations:

1. The arrangements can be alternated. So we only need 2 sets of combinations whose edges do not coincide.

Solution:

1. For odd length:

• one combination can be made by having the first brick as 3X1 and subsequent bricks as 2X1.
• the second combination can be the reverse of the first combination(2X1 bricks followed by one 3X1 brick).

2. For even length:

• one combination can be made by filling with all the 2X1 bricks
• the other combination can be made by having the first brick as 3X1 followed by 2X1 bricks and followed by one last brick of 3X1.

Analysis of Solution:

1. For odd length:

• Total number of 3X1 bricks used= 1
• Total number of 2X1 bricks used = (l-3)/2
• Total bricks in the first and second combinations are the same.

2. For even length:

• Total number of 3X1 bricks in the first combination=0
• Total number of 2X1 bricks in the first combination = l/2
• Total number of bricks in the first combination=l/2
• Total number of 3X1 bricks in the second combination=2
• Total number of 2X1 bricks in the second combination = (l-6)/2
• Total number of bricks in the second combination=2+((l-6)/2)

Edge Cases:

Some of the combinations are not achievable for this kind of arrangement:

1. if the length is 0.
2. if the length is 1.
3. if the length is 4.

Code

Thanks for Reading!

Discord link: https://discord.gg/zEHhKWQm for any questions, debugging sessions.

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