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copied!<p>In many dynamic programming problems, you will build up a 2D table row by row where each row only depends on the row that immediately precedes it. In the case of the 0/1 knapsack problem, the recurrence (from Wikipedia) is the following:</p> <blockquote> <p>m[i, w] = m[i - 1, w] if w<sub>i</sub> > w</p> <p>m[i, w] = max(m[i - 1, w], m[i - 1, w - w<sub>i</sub>] + v<sub>i</sub>) otherwise</p> </blockquote> <p>Notice that all reads from the table when filling row i only come from row i - 1; the earlier rows in the table aren't actually used. Consequently, you could save space in the 2D table by only storing two rows - the immediately previous row and the row you're filling in. You can further optimize this down to just one row by being a bit more clever about how you fill in the table entries. This reduces the space usage from O(nW) (O(n) rows and O(W) columns) to O(W) (one or two rows and O(W) columns).</p> <p>This comes at a cost, though. Many DP algorithms don't explicitly compute solutions as they go, but instead fill in the table, then do a second pass over the table at the end to recover the optimal answer. If you only store one row, then you will get the <em>value</em> of the optimal answer, but you might not know what that optimal answer happens to be. In this case, you could read off the maximum value that you can fit into the knapsack, but you won't necessarily be able to recover what you're supposed to do in order to achieve that value.</p> <p>Hope this helps!</p>

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