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    <p>Think of your pyramid as a tree with the root at the top of the pyramid: I think you want the route with the maximum cost from the root to any of the leaf nodes (bottom of the pyramid). OK, it's not actually a tree, because a node may have two parents, in fact you can get to a node at level <code>i</code> from at most two nodes at level <code>i-1</code>.</p> <p>Anyway, I think you can compute the route with the maximum cost by using dynamic programming. Let me rewrite your data in a matrix like way:</p> <pre><code>7 4 8 1 8 9 2 4 6 7 4 6 7 4 9 4 9 7 3 8 8 </code></pre> <p>and let the missing elements of the matrix be 0. Let's call this matrix <code>v</code> (for values). Now you can build a matrix <code>c</code> (for costs) where <code>c(i,j)</code> is the maximum cost for reaching the tree node at position <code>(i,j)</code>. You can compute it with this recurrence:</p> <p><code>c(i,j) = v(i,j) + max{ c(i-1,j-1), c(i-1,j) }</code></p> <p>where <code>c(h,k)</code> is 0 when you get to a position out of the matrix. Essentially we say that the maximum cost for getting to node at position <code>(i,j)</code> is the cost of the node itself plus the maximum between the costs of getting to its two possible parents at level <code>i-1</code>.</p> <p>Here is <code>c</code> for your example:</p> <pre><code>7 11 15 12 23 24 14 27 30 31 18 33 37 35 40 22 42 44 40 48 48 </code></pre> <p>For instance, let's take <code>i=3, j=2</code>: </p> <pre><code>c(3,2) = v(3,2) + max{ c(2,1), c(2,2) } = 6 + max{ 23 , 24 } = 30 </code></pre> <p>From <code>c</code> you see that the most expensive rout costs 48 (and you have two of them).</p>
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    1. USMarcoS
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    1. COnice job man! I did write the exact same solution except the matrix part, but thanx for your time :) dynamic programming the best!
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      1. USMikee
    2. COyou're welcome! I had fun at working it out :)
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