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    <p>Perhaps it helps if you look at the problem like this:</p> <p>For your example, you have a string like this:</p> <pre><code>AA BB C DDDD E FFF </code></pre> <p>The space positions are the places where you could start a new column. Everywhere else you mustn't to keep same letters in the same column. So you actually can mark the space position like this:</p> <pre><code>AA1BB2C3DDDD4E5FFF </code></pre> <p>Now you have 5 positions where you can either break the column or not, as it's a binary decision, use a string of 0's and 1's for this and brute force every possible combination:</p> <pre><code>12345 00000 -&gt; no break at all, column count = 1, max. lines = 13 ... 01010 -&gt; your example, column count = 3, max. lines = 5 ... 11111 -&gt; breaks everywhere, column count = 6, max. lines = 4 </code></pre> <p>This is a brute force attempt, but you can easily see that the 1's count affects the column count (column count = number of 1's + 1) and you want to minimize max. lines, this should be possible to somehow calculate without having to test each combination.</p> <p>EDIT2: Didn't recognize you want 3 columns, this makes it easier as you know you will have only 3 1's, but it's still brute force.</p> <p>EDIT: Another approach I'd favorize:</p> <p>Write the letter counts like this:</p> <pre><code>A B C D E F 2 2 1 4 1 3 </code></pre> <p>You can now join letters that are next to each other. Always chose the two letters with the lowest count sum:</p> <pre><code>2 2 1 4 1 3 - lowest = "2 1" 2 3 4 1 3 - lowest = "1 3" 2 3 4 4 - lowest = "2 3" 5 4 4 - stop now, as we have 3 columns now Result: AABBC, DDDD, EFFF </code></pre> <p>This perhaps won't lead to the optimal solution, but it's a nice and easy way to solve your problem, I think.</p>
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    1. COYour second solution looks beautiful on paper but I'm not sure how I'd do that programmatically. Looks like there'd be an ungodly amount of repetition.
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    2. COYou could easily do it programmatically. Use a list or an array that contains the letter counts at first. Then calculate sum(i)=count(i)+count(i+1) for each item and join where sum(i) is lowest. Do this until you have only 3 columns left.
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    3. COOne problem with your (not terribly unattractive) greedy solution is that you can't handle tiebreaks. Consider the case where you only want two breaks and the sequence is 2 1 2 2. The optimal division is 3/4, but if you group 1 to the right you can't achieve it.
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