Combine the previous three versions of wc. Given a list of lines and a mode flag, return the requested count(s) as a string. This is our own simplified interface, not the real wc flags: the real tool combines flags (such as -lw) and pads its columns with spaces.
Function signature:
def wc(self, lines: List[str], mode: str) -> str:
Mode values:
"l"→ line count, as a string."w"→ word count (whitespace-separated tokens), as a string."c"→ character count (lines only, no separating newlines), as a string."all"→lines,words,chars(in that order) joined with a tab character:f"{lines}\t{words}\t{chars}".
The "c" count is the one from v3 (the sum of the line lengths, no newlines). Any other mode value, for example "x" or "L" (modes are case-sensitive), gives the same result as "all". For an empty list, "l", "w" and "c" give "0" and "all" gives "0<TAB>0<TAB>0". The input is plain ASCII text.
Example:
Input: lines = ["hello", "world"], mode = "all" Output: "2\t2\t10"
2 lines, 2 words (one per line), 10 characters total.
Compute all three counts upfront — even if you only return one. Then a simple chain of
if statements selects which count(s) to return.
The function always returns a string — so
str(numLines) for the single-mode cases, and an f-string for "all".
numLines = len(lines)
numWords = sum(len(line.split()) for line in lines)
numChars = sum(len(line) for line in lines)
The full solution:
class Solution:
def wc(self, lines, mode):
numLines = len(lines)
numWords = sum(len(line.split()) for line in lines)
numChars = sum(len(line) for line in lines)
if mode == "l":
return str(numLines)
if mode == "w":
return str(numWords)
if mode == "c":
return str(numChars)
return f"{numLines}\t{numWords}\t{numChars}"
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