Java > String-3 > sameEnds
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Given a string, return the longest substring that appears at both the beginning and end of the string without overlapping. For example, sameEnds("abXab") is "ab". sameEnds("abXYab") → "ab" sameEnds("xx") → "x" sameEnds("xxx") → "x" ...Save, Compile, Run public String sameEnds(String string) { } See also Java Example Code. Java help docs: If Boolean Logic | Strings | While and For Loops | Arrays and Loops prev  |  next  |  chance   |  CodingBat  >  String-3

Forget It! -- delete my code for this problem

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