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# lot of junk in the sequence, the number of *unique* junk
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# elements is probably small. So the memory burden of keeping
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# this dict alive is likely trivial compared to the size of b2j.
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def find_longest_match(self, alo, ahi, blo, bhi):
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"""Find longest matching block in a[alo:ahi] and b[blo:bhi].