```My solution finds substrings of length n, groups the identical ones together in a hash of their starting indices, prunes out cases where two of the "repeated" substring overlap each other, and then looks in the same places for substrings of length n+1.

The aggressive pruning does, however, cause it to fail in the case where a substring that overlaps at one length does not overlap at a higher length. For example, take the string "aaaaaa". The substring "aa" exists at indicies [0,1,2,3,4]. My program will prune that down to [0,2,4]. It will then record the substring of length 3 that starts at each of [0,2,4], prune out the ones that start at 2 (overlaps with the one startign at 0) and 4 (goes out of bounds), leaving only an unrepeated substring, and would thus fall back and print "aa," even though the correct answer is "aaa," because the index of 3 had already been pruned.

On the other hand, the aggressive pruning does make my program somewhat fast. It takes about 45s on Also Sprach Zarathustra.

substr_locs = {}

text.split(//).each_with_index do |c,idx|
if substr_locs[c]
substr_locs[c] << idx
else
substr_locs[c] = [idx]
end
end

substr_locs.each_pair do |substr,locs|
substr_locs.delete(substr) if locs.length == 1
end

len = 1
done = false
until done
len += 1
new_substr_locs = {}
substr_locs.each_pair do |substr,locs|
locs.each do |idx|
s = text[idx,len]
next if idx+len>text.length
if new_substr_locs[s]
new_substr_locs[s] << idx
else
new_substr_locs[s] = [idx]
end
end
end
##Must reduce for the case where a substring overlaps itself
new_substr_locs.each_key do |s|
locs = new_substr_locs[s]
idx = 1
while idx < locs.length
if locs[idx]-locs[idx-1] < len
locs.delete_at(idx)
else
idx += 1
end
end
new_substr_locs.delete(s) if locs.length == 1
end
unless new_substr_locs.keys.empty?
substr_locs = new_substr_locs
else
done = true
end
end
puts substr_locs.keys.first

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```