Squash Spaces - codepath/compsci_guides GitHub Wiki
Unit 4 Session 1 (Click for link to problem statements)
TIP102 Unit 1 Session 2 Advanced (Click for link to problem statements)
Problem Highlights
- 💡 Difficulty: Easy
- ⏰ Time to complete: 10 mins
- 🛠️ Topics: Strings, Two-Pointer Technique
U-nderstand
Understand what the interviewer is asking for by using test cases and questions about the problem.
-
Q: What is the input to the function?
- A: The input is a string
s
that may contain spaces between words, including leading and trailing spaces.
- A: The input is a string
-
Q: What is the expected output of the function?
- A: The function should return a new string where consecutive spaces are reduced to a single space, and the string is trimmed of any leading or trailing spaces.
-
Q: How should the function handle a string with no spaces?
- A: If there are no spaces in the string, the function should return the original string.
-
Q: What should the function return if the string contains only spaces?
- A: The function should return an empty string.
-
The function
squash_spaces()
should reduce consecutive spaces in the input string to a single space while trimming any leading or trailing spaces.
HAPPY CASE
Input: " Up, up, and away! "
Expected Output: "Up, up, and away!"
UNHAPPY CASE
Input: "With great power comes great responsibility."
Expected Output: "With great power comes great responsibility."
EDGE CASE
Input: " "
Expected Output: "
P-lan
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Use two pointers to traverse the string, appending characters to the result list while skipping consecutive spaces.
1. Initialize a pointer `ptr` to 0 (this will read through the input string).
2. Create an empty list `output` to store the result characters.
3. Skip initial spaces:
a. While `ptr` is less than the length of the string `s` and `s[ptr]` is a space:
i. Increment `ptr`.
4. Iterate through the string `s` while `ptr` is less than the length of `s`:
a. If the current character is not a space:
i. Append `s[ptr]` to `output`.
b. If the current character is a space:
i. Check if `output` is not empty and the last character in `output` is not a space.
ii. Also, ensure there is a non-space character following the current space.
iii. If both conditions are met, append a single space to `output`.
c. Increment `ptr` by 1.
5. Convert the `output` list to a string:
a. Join the characters in the `output` list to form the final result string.
6. Return the final result string.
⚠️ Common Mistakes
- Forgetting to check for boundaries when accessing string indices.
- Not handling leading or trailing spaces properly.
I-mplement
Implement the code to solve the algorithm.
def squash_spaces(s):
# Initialize pointers and the output list
ptr = 0
output = []
# Skip initial spaces
while ptr < len(s) and s[ptr] == ' ':
ptr += 1
while ptr < len(s):
if s[ptr] != ' ':
# Add non-space characters directly to output
output.append(s[ptr])
else:
# Add a space only if the last added character is not a space
# and there are more characters after the current one
if len(output) > 0 and output[-1] != ' ' and ptr + 1 < len(s) and s[ptr + 1] != ' ':
output.append(s[ptr])
ptr += 1
# Join list into a final string
return ''.join(output)