Strings

In this module

Find how many characters a string holds with len().
Read one character with indexing (word[i]), including from the end with a negative index.
Take out part of a string with slicing (word[a:b]).
Convert between a character and its code point with ord() and chr().

Try these on the judge. Each link opens the problem on DMOJ.

2015 J2
Happy or Sad (opens on DMOJ in a new tab) DMOJ
Count two kinds of faces in a line of text and decide whether it reads happy, sad, unsure, or neither.
Why DMOJ: Tries the same kind of problem on DMOJ, which holds 2014 to 2020.
2019 J2
Time to Decompress (opens on DMOJ in a new tab) DMOJ
Turn each "count and character" line into that character repeated that many times.
Why DMOJ: Tries the same kind of problem on DMOJ, which holds 2014 to 2020.
2018 J2
Occupy parking (opens on DMOJ in a new tab) DMOJ
Compare two lines and count where both mark the same spot.
Why DMOJ: Tries the same kind of problem on DMOJ, which holds 2014 to 2020.