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copied!<p>I think the solution is pretty easy.</p> <p>Start with an array <code>x</code> initialised to <code>empty</code> values such that there is one space for each item you need to place.</p> <p>Then, for each <code>(item, frequency)</code> pair in descending order of frequency, assign <code>item</code> values to <code>x</code> in alternating slots starting from the first <code>empty</code> slot.</p> <p>Here's how it works for your example:</p> <pre><code>20*A A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A 10*B ABABABABABABABABABABA_A_A_A_A_A_A_A_A_A 5*C ABABABABABABABABABABACACACACACA_A_A_A_A 2*E ABABABABABABABABABABACACACACACAEAEA_A_A 1*F ABABABABABABABABABABACACACACACAEAEAFA_A </code></pre> <p>At this point we fail, since <code>x</code> still has an empty slot. Note that we could have identified this right from the start since we need at least 19 slots between the <code>A</code>s, but we only have 18 other items.</p> <p><strong>UPDATE</strong></p> <p>Leonidas has now explained that the items should be distributed "evenly" (that is, if we have k items of a particular kind, and n slots to fill, each "bucket" of n/k slots must contain one item of that kind.</p> <p>We can adapt to this constraint by spreading out our allocations rather than simply going for alternating slots. In this case (and let's assume 2 Fs so we can solve this), we would have</p> <pre><code>20*A A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A_A 10*B ABA_ABA_ABA_ABA_ABA_ABA_ABA_ABA_ABA_ABA 5*C ABACABA_ABACABA_ABACABA_ABACABA_ABACABA 2*E ABACABAEABACABA_ABACABAEABACABA_ABACABA 2*F ABACABAEABACABAFABACABAEABACABAFABACABA </code></pre>

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