How should I practice BitFont Part 3 - Decode Run-Length-Encoded Rows?

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Problem · String

BitFont Part 3 - Decode Run-Length-Encoded Rows

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This continues the bitmap-font renderer. To save space, a font may store each glyph row Run-Length-Encoded (RLE). Decode the encoded rows back into full 0/1 strings.

Each encoded row is a sequence of run-length characters. The runs alternate pixel state and always start with off (0): the first run is off, the next on, the next off, and so on. Each run-length character maps to a count:

digits 0-9 mean lengths 0 through 9;
letters a-z mean lengths 10 through 35 (so a = 10, b = 11, ..., z = 35).

For each run character, emit that many copies of the current pixel value, then flip the state for the next run. A run of length 0 emits no pixels but still flips the state — do not special-case zero by skipping the flip, or every later run in that row inverts. All decoded rows end up the same width.

Examples

01 · Example 1

encodedRows = ["532", "19"]
return = ["0000011100", "0111111111"]

Row '532': 5 off (00000), 3 on (111), 2 off (00) gives 0000011100. Row '19': 1 off (0), 9 on (111111111) gives 0111111111. Both rows are width 10.

02 · Example 2

encodedRows = ["05", "a0"]
return = ["11111", "0000000000"]

Row '05': the first run length is 0 (off) which emits nothing but still flips to on, then 5 on gives 11111. Row 'a0': a = 10 off gives ten 0s, then a 0-length on run emits nothing. This exercises the zero-length-run flip trap.

03 · Example 3

encodedRows = ["3b"]
return = ["00011111111111"]

Row '3b': 3 off (000), then b = 11 on (11111111111) gives 00011111111111, a width of 14.

Constraints

Each encoded row is a string of run-length characters in 0-9 or a-z.
Runs alternate state starting from off (0): off, on, off, on, ...
0-9 map to lengths 0-9; a-z map to lengths 10-35.
A length-0 run emits nothing but still flips the state.
All decoded rows have the same width.