String-3 > sameEnds
prev  |  next  |  chance

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" Go...Save, Compile, Run (ctrl-enter) public String sameEnds(String string) { } Go Editor font size %:75100125150 Shorter output

Forget It! -- delete my code for this problem

Progress graphs:
 Your progress graph for this problem
 Random user progress graph for this problem
 Random Epic Progress Graph

Java Help

• Java Example Solution Code
• Java String Introduction (video)
• Java Substring v2 (video)
• Java String Equals and Loops
• Java String indexOf and Parsing
• Java If and Boolean Logic
•   If Boolean Logic Example Solution Code 1 (video)
•   If Boolean Logic Example Solution Code 2 (video)
• Java For and While Loops
• Java Arrays and Loops
• Java Map Introduction
• Java Map WordCount
• Java Functional Mapping
• Java Functional Filtering

Misc Code Practice

• Code Badges
• Introduction to Mod (video)
• MakeBricks problem and solution (video x 2)
• FizzBuzz the famous code interview question (video)

Difficulty: 321.0

Copyright Nick Parlante 2017 - privacy