Different organisms prefer different codons for the same amino acid, a phenomenon
called codon usage bias. Measuring it matters when designing genes for expression in a host organism —
a human gene expressed in E. coli often fails simply because it uses codons the bacterium reads slowly.
Counting codons is also the natural first use of a dictionary, which is the data structure biology code
reaches for most often.
Statement
Given a coding DNA sequence, split it into non-overlapping codons of three bases,
starting at position 0, and count how many times each codon appears.
Ignore any trailing bases that do not form a complete codon.
Print each distinct codon and its count on its own line, separated by a single space, sorted by count
descending. Where counts tie, sort by codon ascending lexicographically.
Input — read from standard input
| Variable | Type | Description |
|---|---|---|
cds
line 1
|
str |
The coding DNA sequence
3 <= len(cds) <= 3000, uppercase A, C, G, T only
|
These variables are already read for you in the starter code on the right.
Output
str one line per distinct codon, codon and count separated by a space, sorted by count desc then codon asc
Sample Cases
ATGATGGCCTAAATG 2
GCC 1
TAA 1
AAACCCAAA 1
CCC 1
Submit also runs your code against 4 hidden test cases. Hidden inputs are never shown — if one fails you'll get its number and a description of the mismatch, not the data.
Constraints
3 <= length(cds) <= 3000- Codons are read in non-overlapping frames starting at position 0
- Any trailing 1 or 2 bases are ignored
- Sort by count descending, then by codon ascending — this makes output deterministic
Further Reading
range(0, len(cds) - 2, 3)walks the sequence in complete codons only.- A
dictwith.get(codon, 0) + 1accumulates counts without needing to pre-populate keys. - Sort with a key such as
(-count, codon)to get descending counts and ascending ties in one pass.