6 lines
213 B
Haskell
6 lines
213 B
Haskell
leftmostLongest :: (Foldable t) => [t a] -> t a
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leftmostLongest xs =
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let lastElem = (last xs)
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filteredLst = (filter (\val -> (length val) == (length lastElem)) xs)
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in head filteredLst
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