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    <p>It can be solved in O(<em>n</em>), like this:</p> <p>C[<em>i</em>] represents the maximum cash you can hold at the end of day <em>i</em>, and G[<em>i</em>] represents the maximum amount of goods you can hold at the end of day <em>i</em>. I'm going to suppose you start out with 1 unit of cash and no goods, but the algorithm will work for other initial conditions.</p> <ul> <li>Initialize: C[0] = 1; G[0] = 0. </li> <li>Repeat for <em>i</em> from 1 to <em>n</em>:</li> <li>Let the selling price on day <em>i</em> be S[<em>i</em>] and the buying price B[<em>i</em>].</li> <li>Set C[<em>i</em>] = max(C[<em>i</em> − 1], G[<em>i</em> − 1]S[<em>i</em>]) and G[<em>i</em>] = max(G[<em>i</em> − 1], C[<em>i</em> − 1]/B[<em>i</em>]).</li> </ul> <p>Here's an implementation in Python:</p> <pre class="lang-py prettyprint-override"><code>from __future__ import division def max_cash(cash, goods, selling_price, buying_price): """Return the maximum amount of cash you can have, if you start with the given cash and goods, and on day i you can sell goods at selling_price[i] or buy them at buying_price[i]. Cash and goods are assumed to be arbitrarily divisible. """ for sell, buy in zip(selling_price, buying_price): assert(sell &lt;= buy) cash, goods = max(cash, goods * sell), max(goods, cash / buy) return cash </code></pre> <p>Let's check this on some of the alleged counter-examples raised in comments:</p> <ul> <li><p><a href="https://stackoverflow.com/questions/4128487/solving-buying-selling-optimization-problem/4128648#comment4463612_4128648">JPvdMerwe</a> suggests that this case "breaks all greedy algorithms I can think of", but the algorithm above works fine:</p> <pre><code>&gt;&gt;&gt; max_cash(0, 1, [2 ** i for i in range(10)], [float('inf')] * 10) 512 </code></pre></li> <li><p><a href="https://stackoverflow.com/questions/4128487/solving-buying-selling-optimization-problem/4128648#comment4464004_4128648">Keegan Carruthers-Smith</a> suggests this counter-example:</p> <pre><code>&gt;&gt;&gt; max_cash(1, 0, [2, 2, 2], [2, 2, 2]) 1 </code></pre> <p>which is unsurprising since the selling price is always the same as the buying price, and so no profit can be made, and so you can't possibly end up with more cash than you started with.</p></li> </ul>
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    1. POSolving buying-selling optimization problem
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    1. USGareth Rees
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    1. COMaybe you should just make it explicit, that each day you can either do one of 3 things: sell everything, buy as much as possible or do nothing. Then you also need to say that it doesn't make sense to sell only a part of your stock or buy only using some of your money. Buying and selling on the same day is pointless as: S[i] ≤ B[i] . Then you just need to give a reasonable argument why maximising money at the end of the day maximises profits at the end. Since this is a greedy algorithm.
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    2. COI'm leaving proof of correctness as an exercise for the reader.
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    3. COOk, after some more thought, I've realised the last bullet point is completely wrong. When selling you dont keep money you had on the previous day. And so on
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