Unit 4 Session 1 (Click for link to problem statements)

TIP102 Unit 1 Session 2 Advanced (Click for link to problem statements)

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Strings, Two-Pointer Technique

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.

• 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: " 

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.

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)