Balanced layout of n items in a grid.

Question

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

Answer 1

• 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

Answer 2

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]

Answer 3

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

Answer 4

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