CODEWITHSUNDEEP
algorithmmath
Riyas
Riyas

Reputation: 327

Balanced layout of n items in a grid

I have a list of n logos to display in a grid, with a maximum of 3 per row. What's an algorithm to decide how many to display per row such that the number of logos per row is as balanced as possible without using more than the minimum possible number of rows?

For example:

 n -> number in each row
 1 -> 1
 2 -> 2
 3 -> 3
 4 -> 2, 2
 5 -> 3, 2
 6 -> 3, 3
 7 -> 3, 2, 2
 8 -> 3, 3, 2
 9 -> 3, 3, 3
10 -> 3, 3, 2, 2

Upvotes: 3

Views: 975

Answers (4)

anon
anon

Reputation:

just use n/3 to calculate the row and n%3 to calculate the column

edit: ok i saw you edited your question.... i din't saw that you want to display 2 in each row if the are 4 logos. but then you can use n mod 3 to calculate if their is a reminder as others already suggested

if n%3 = 0 then just put 3 logos in each row if n%3 = 1 then put the last 4 logos in two rows if n%3 = 2 then put 3 logos in n row and the last 2 logos in a separate row

Upvotes: 1

MikeyB
MikeyB

Reputation: 3360

AS confusing as your question is, I think what you need to do is first determine:

number_of_rows = ceil(number_of_logos / 3.0)

Then add a logo to each row, one at a time.

Python:

import math
def partition_logos(count, lsize):
    num_lines = int(math.ceil(count / float(lsize)))
    partition = [0] * num_lines
    for i in xrange(count):
        partition[i%num_lines] += 1
    return partition

>>> for i in xrange(1,11):
...     print partition_logos(i, 3)
[1]
[2]
[3]
[2, 2]
[3, 2]
[3, 3]
[3, 2, 2]
[3, 3, 2]
[3, 3, 3]
[3, 3, 2, 2]

Upvotes: 6

Mark Byers
Mark Byers

Reputation: 839044

  • For N <= 3 just use N.
  • If N is exactly divisible by 3 then use: 3 3 ... 3
  • If N when divided by 3 has remainder 1 then use: 3 3 ... 2 2
  • If N when divided by 3 has remainder 2 then use: 3 3 ... 3 2

Upvotes: 12

Heath Hunnicutt
Heath Hunnicutt

Reputation: 19467

A recursive solution, in Python:

def logos_by_row(N, rows):
    width = 0
    if N > 4 or N == 3:
        width = 3
    elif N == 4 or N == 2:
        width = 2
    elif N == 1:
        width = 1

    if width != 0:
        rows.append(width)
        logos_by_row(N - width, rows)


answer = []
for i in range(10):
    logos_by_row(i+1, answer)
print answer

Upvotes: 1

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