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    <p>So the OP's problem in general can be solved by the DP algorithm given in <a href="https://stackoverflow.com/questions/6454598/stuck-with-an-interview-question-partitioning-of-an-array">this question</a>. I'll repeat the crux of it here for completeness:</p> <blockquote> <p>Suppose <em>d[i][j]</em> is solution of the problem when we have items <em>s[1], .., s[i]</em> and <em>j</em> containers. Then:</p> <ol> <li><em>d[0][j] = 0</em> for each <em>j</em></li> <li><em>d[i][1] = Sum(s[1], ..., s[i])</em> for each <em>i</em></li> <li><em>d[i][j] = min(max(d[i-t][j-1], Sum(s[i-t+1], ..., s[i]) over all 1&lt;=t&lt;=i)</em></li> </ol> </blockquote> <p>However, the naive implementation consumes O(NK) space, which is too much. Fortunately, look at (3): <code>d[_][j]</code>'s value depends only on <code>d[_][j-1]</code>. This means that once we're done computing all <code>d[_][j]</code>, we can use the space we were using to store <code>d[_][j-1]</code> to instead store the values of <code>d[_][j+1]</code> we're about to compute.</p> <p>Since the values <code>d[0][j]</code> are all zero, we don't need to store those either. Hence the only arrays we need to store are the O(N)-sized <code>d[i][1]</code>, the O(N)-sized array that holds the result from the <code>j-1</code>th iteration, and the O(N)-sized array that holds the results from the <code>j</code>th iteration we're currently computing. Total: O(N).</p> <p>Edit: So first time round, I didn't actually answer the OP's question about how to count the number of containers at max weight. Suppose <code>w[i][j]</code> holds the max weight of the <code>i,j</code>th problem, and <code>c[i][j]</code> is the number of containers which match that max weight. Then:</p> <ol> <li><code>c[0][j] = j</code> and <code>w[0][j] = 0</code> for each <code>j</code></li> <li><code>c[i][1] = 1</code> and <code>w[i][1] = Sum(s[1], .., s[i])</code> for each <code>i</code></li> <li>Let <code>u</code> be equal to the <code>t</code> chosen in pt (3) above. <ul> <li>If <code>d[i-u][j-1] &gt; Sum(s[i-u+1], .., s[i])</code>: <ul> <li>then <code>c[i][j] = c[i-u][j-1]</code> and <code>w[i][j] = w[i-u][j-1]</code></li> <li>because the <code>j</code>th container with weight <code>Sum(s[i-u+1], .., s[i]</code> is lighter than the heaviest container out of containers <code>1, .., j-1</code>, which has weight <code>d[i-u][j-1]</code>. </li> </ul></li> <li>If <code>d[i-u][j-1] &lt; Sum(s[i-u+1], .., s[i])</code>: <ul> <li>then <code>c[i][j] = 1</code> and <code>w[i][j] = Sum(s[i-u+1], .., s[i])</code> </li> <li>because the <code>j</code>th container with items <code>s[i-u+1], .., s[i]</code> is the new heaviest container.</li> </ul></li> <li>If If <code>d[i-u][j-1] == Sum(s[i-u+1], .., s[i])</code>: <ul> <li>then <code>c[i][j] = c[i-u][j-1]+1</code> and <code>w[i][j] = d[i-u][j-1]</code></li> <li>because the <code>j</code>th container is the same weight as the previous heaviest container.</li> </ul></li> </ul></li> </ol> <p>Again, we only need to use a constant number of <code>O(N)</code> sized arrays, for <code>O(N)</code> time overall.</p>
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    1. COThanks for your reply. As I understand, this algorithm provide us information about maximum weight. How can I get information about number of containers with such weight? Or am I missing something?
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    2. COCompletely forgot about that, sorry. Edited a solution in.
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