CF 291B - Command Line Arguments

Rating: 1300
Tags: *special, implementation, strings
Solve time: 1m 26s
Verified: yes

Solution

Problem Understanding

We are given a single string representing a command line in a simplified Pindows operating system. Each "lexeme" in this command line is either a contiguous sequence of non-space characters or a sequence of characters enclosed in double quotes. The first lexeme is the program name, and all subsequent lexemes are its arguments. The goal is to parse this string into individual lexemes and print them in order, surrounded by angle brackets.

The input string can include uppercase and lowercase letters, digits, punctuation marks like ., ,, ?, !, quotes, and spaces. Quotes are used only to group lexemes that contain spaces or to represent empty strings. Every opening quote has a matching closing quote, and quotes cannot be nested. There may be spaces outside quotes that separate lexemes, but spaces inside quotes are part of the lexeme.

The constraints allow strings up to length 100,000. This implies we need an algorithm that works in linear time; any solution that inspects characters multiple times or performs repeated string concatenations inefficiently would risk exceeding the time limit. A naive split-on-spaces approach fails because it cannot handle quotes and empty strings correctly. Similarly, blindly scanning for quotes without careful management of boundaries may merge lexemes incorrectly.

Non-obvious edge cases include strings with empty lexemes (""), lexemes that consist only of spaces (" "), or strings that start or end with quotes. For example, the input "a" "" " " should produce <a>, <>, < >, not <a> < > < > or similar errors. A careless parser could skip empty strings or misinterpret spaces outside quotes as part of a lexeme.

Approaches

The brute-force approach would be to iterate over the string, attempting to split on spaces unless we are inside quotes. This works correctly if implemented carefully because the string is guaranteed to be valid and all quotes are balanced. Each character must be inspected to decide whether it is part of a lexeme or a separator, so the operation count is O(n). Any repeated string slicing or concatenation could push this to O(n^2) in the worst case because each concatenation copies the substring.

The key insight for a more efficient and robust solution is to scan the string character by character while maintaining a flag that indicates whether we are inside a quoted lexeme. When we see a quote, we toggle this flag. While inside quotes, all characters, including spaces, are added to the current lexeme. Outside quotes, spaces indicate the end of a lexeme, and non-space characters start a new lexeme. This linear scan guarantees O(n) time without extra overhead and naturally handles empty strings and spaces inside quotes.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(n)	Risky due to string concatenation cost
Linear Scan with Quote Flag	O(n)	O(n)	Accepted

Algorithm Walkthrough

Initialize an empty list to store the lexemes and an empty string to build the current lexeme. Also initialize a boolean flag in_quotes to False.
Iterate through each character of the input string.
If the current character is a double quote, toggle the in_quotes flag. If toggling in_quotes from True to False (i.e., closing a quote), append the current lexeme to the list and reset the current lexeme to empty.
If in_quotes is True and the character is not a quote, add it to the current lexeme. This ensures that all spaces and punctuation inside quotes are preserved.
If in_quotes is False, spaces indicate separation between lexemes. When a space is encountered, if the current lexeme is non-empty, append it to the list and reset the current lexeme.
For any non-space character outside quotes, append it to the current lexeme. This collects normal unquoted lexemes.
After finishing the iteration, if the current lexeme is non-empty, append it to the list. This handles the last lexeme if the string does not end with a space or quote.
Print each lexeme surrounded by angle brackets.

The invariant throughout the loop is that current_lexeme always contains the characters of the lexeme currently being read, and in_quotes accurately reflects whether we are inside a quoted block. This guarantees that all characters inside quotes are treated as part of a single lexeme and that spaces outside quotes split lexemes correctly.

Python Solution

import sys
input = sys.stdin.readline

s = input().rstrip()
lexemes = []
current_lexeme = []
in_quotes = False
i = 0
n = len(s)

while i < n:
    c = s[i]
    if c == '"':
        in_quotes = not in_quotes
        if not in_quotes:
            lexemes.append(''.join(current_lexeme))
            current_lexeme = []
    elif in_quotes:
        current_lexeme.append(c)
    else:
        if c == ' ':
            if current_lexeme:
                lexemes.append(''.join(current_lexeme))
                current_lexeme = []
        else:
            current_lexeme.append(c)
    i += 1

if current_lexeme:
    lexemes.append(''.join(current_lexeme))

for lex in lexemes:
    print(f"<{lex}>")

This solution carefully toggles the in_quotes flag on each quote and builds lexemes character by character. It handles empty lexemes ("") correctly by appending an empty string when a quote closes immediately. Spaces outside quotes terminate lexemes, while spaces inside quotes are preserved. The final check ensures that a lexeme at the end of the string is not omitted.

Worked Examples

Example 1

Input: "RUn.exe O" "" " 2ne, " two! . " "

i	c	in_quotes	current_lexeme	lexemes
0	"	False→True	[]	[]
1-9	R U n . e x e O	True	['R','U','n','.','e','x','e',' ','O']	[]
10	"	True→False	[]	['RUn.exe O']
11	space	False	[]	['RUn.exe O']
12	"	False→True	[]	['RUn.exe O']
13	"	True→False	[]	['RUn.exe O','']
14	space	False	[]	['RUn.exe O','']
15	"	False→True	[]	['RUn.exe O','']
16-22	space 2 n e , space	True	[' ',' ',' ','2','n','e',',',' ']	['RUn.exe O','']
23	"	True→False	[]	['RUn.exe O','', ' 2ne, ']
24	space	False	[]	['RUn.exe O','', ' 2ne, ']
25-28	t w o !	False	['t','w','o','!']	['RUn.exe O','', ' 2ne, ']
29	space	False	[]	['RUn.exe O','', ' 2ne, ','two!']
30	.	False	['.']	['RUn.exe O','', ' 2ne, ','two!']
31	space	False	[]	['RUn.exe O','', ' 2ne, ','two!','.']
32	"	False→True	[]	['RUn.exe O','', ' 2ne, ','two!','.']
33	space	True	[' ']	['RUn.exe O','', ' 2ne, ','two!','.']
34	"	True→False	[]	['RUn.exe O','', ' 2ne, ','two!','.',' ']

Output matches the sample.

Example 2

Input: abc " " def

i	c	in_quotes	current_lexeme	lexemes
0-2	a b c	False	['a','b','c']	[]
3	space	False	[]	['abc']
4	"	False→True	[]	['abc']
5	space	True	[' ']	['abc']
6	"	True→False	[]	['abc',' ']
7	space	False	[]	['abc',' ']
8-10	d e f	False	['d','e','f']	['abc',' ']
end			[]	['abc',' ','def']

Output:

<abc>
< >
<def>

This trace demonstrates correct handling of spaces inside quotes and proper lexeme separation outside quotes.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each character is processed exactly once, with appending to a list of characters being O(1) per character
Space	O(n)	We store all lexemes and build each lexeme in a list of characters

For n