Re^2: Challenge: Fast Common Substrings

[blokhead]

++ Wow, thank you for introducing me to suffix trees. What an interesting concept, and how refreshing to see a linear-time algorithm for constructing such a creature. I see you've used the javascript applet at this page, which others may want to check out.

However, I'd like to slightly revise the algorithm you outlined. Consider the following example:

string = ababc%bc$

     |      |(3:abc%bc$)|leaf
     |(1:ab)|
     |      |(5:c%bc$)|leaf
tree:|
     |     |(3:abc%bc$)|leaf
     |(2:b)|
     |     |     |(6:%bc$)|leaf
     |     |(5:c)|
     |     |     |(9:$)|leaf
     |
     |     |(6:%bc$)|leaf
     |(5:c)|
     |     |(9:$)|leaf
     |
     |(6:%bc$)|leaf
     |
     |(9:$)|leaf

"ab" appears twice in the first string, and so it gives a node with two leaves. The actual condition you should check is whether a node has one leaf containing the % separator and another leaf without the % symbol.