In this post, we’ll dive into LeetCode Problem #2109: Adding Spaces to a String. This problem involves manipulating strings based on a set of indices, making it a fantastic opportunity to brush up on string slicing and iteration. We’ll explore the problem step-by-step, tackle it with an optimized solution, and showcase an example walkthrough using higher numbers for clarity. Let’s get started! 🌟

💡 Problem Statement

You are given a string s and an array of integers spaces, where each integer represents an index in the string s. Your task is to insert a single space at each index in the spaces array. Return the resulting string after adding all the spaces.

🔖 Constraints:

• 1 <= s.length <= 3 * 10⁵
• 1 <= spaces.length <= 3 * 10⁵
• 0 <= spaces[i] < s.length
• All values in spaces are strictly increasing.

🌟 Example

Input:

s = "LeetCodeHelpsYouLearn"
spaces = [8, 13, 18]

Output:

"LeetCode Helps You Learn"

Explanation:

• Insert a space at index 8: "LeetCode HelpsYouLearn"
• Insert a space at index 13: "LeetCode Helps YouLearn"
• Insert a space at index 18: "LeetCode Helps You Learn"

🛠️ Basic Solution 💻

Let’s dive deeper into the basic solution for solving the problem. This section will explore the underlying mechanics, alternative approaches, and potential pitfalls of this solution, providing a thorough understanding.

🔍 Step-by-Step Breakdown

1. Initialization:
   • The function starts by initializing two variables:
     • index to 0: This serves as a pointer to track the current position in the string s.
     • result as an empty list: This will store the substrings of s between consecutive indices in spaces.
2. Iterating Through spaces:
   • For each index in the spaces array, the string is sliced from the index pointer to the current space index:
     • For example, if s = "LeetCodeHelpsYouLearn" and spaces = [8, 13, 18], in the first iteration, s[0:8] results in "LeetCode".
   • The sliced substring is added to result, and the index pointer is updated to the current space index.
3. Appending the Remaining Substring:
   • After all indices in spaces have been processed, the substring from the last index in spaces to the end of the string is appended to result.
   • For instance, in the example above, the final substring s[18:] results in "Learn".
4. Joining Substrings:
   • The join() method concatenates all substrings in result with spaces in between, forming the desired output.

🔑 Why This Works

1. String Slicing:
   • Python string slicing is efficient because it operates on the original string without creating a new copy for each slice.
   • This ensures that slicing operations are performed in (O(1)) time for each slice.
2. List and Join Operations:
   • Appending substrings to a list (result) and joining them at the end ensures that string concatenation, which can be costly if done repeatedly in a loop, is performed only once.
3. Strictly Increasing Indices:
   • The problem guarantees that indices in spaces are strictly increasing. This simplifies the slicing process, as we don’t need to handle overlapping or unordered indices.

🛑 Common Pitfalls

1. Incorrect Slicing:
   • If you forget to update the index pointer after slicing, the program will repeatedly slice from the same starting point, producing incorrect results.
   • Example:

     s = "LeetCodeHelpsYouLearn"
     spaces = [8, 13, 18]
     # Forgetting to update `index` would result in:
     result = [s[0:8], s[0:13], s[0:18]]
     

2. Off-by-One Errors:
   • It’s crucial to ensure that slices are properly aligned. For example, if you slice from index+1 instead of index, the substring would skip one character.
3. Handling Empty Strings or Arrays:
   • Ensure that edge cases like s = "" or spaces = [] are handled gracefully without throwing errors.

🚀 Optimization Considerations

The basic solution is already optimal, but let’s explore why it’s efficient compared to alternative approaches:

Alternative Approach: Direct String Insertion

A naive approach would involve directly inserting spaces into the string s:

for space in spaces:
    s = s[:space] + " " + s[space:]

• Drawbacks:
  • Each insertion creates a new string, resulting in (O(n^2)) time complexity for multiple insertions.
  • For large inputs, this approach becomes infeasible due to high memory usage and slower execution.

Why List-Based Approach is Better:

• By collecting all substrings in a list (result) and joining them at the end, we minimize the number of string concatenation operations, keeping the time complexity at (O(n)).

📝 Detailed Walkthrough of the Code

Let’s walk through the code step-by-step using a simple example:

Input:

s = "CodingIsFun"
spaces = [6, 8]

Execution:

1. Initialization:

   index = 0
   result = []
   

2. First Iteration (space = 6):
   • Slice: s[0:6] → "Coding"
   • Update result: result = ["Coding"]
   • Update index: index = 6
3. Second Iteration (space = 8):
   • Slice: s[6:8] → "Is"
   • Update result: result = ["Coding", "Is"]
   • Update index: index = 8
4. Appending Remaining String:
   • Slice: s[8:] → "Fun"
   • Update result: result = ["Coding", "Is", "Fun"]
5. Joining Substrings:
   • Final Output: " ".join(result) → "Coding Is Fun"

🔎 Insights

1. Efficiency in Handling Large Strings:
   • The list-based approach processes substrings independently, making it well-suited for large inputs where direct insertion would be computationally expensive.
2. Simplicity and Readability:
   • The algorithm is straightforward to understand and implement, leveraging Python’s powerful slicing and list operations.
3. Scalability:
   • Even for the upper constraint (s.length = 3 * 10⁵), the algorithm performs efficiently due to its linear time complexity.

🛠️ Python Code Recap

class Solution:
    def addSpaces(self, s: str, spaces: List[int]) -> str:
        index, result = 0, []
        for space in spaces:
            result.append(s[index:space])
            index = space
        result.append(s[index:])
        return " ".join(result)

This method provides a robust and scalable solution to the problem, leveraging Python’s string and list operations effectively.

⏱️ Time Complexity

• Slicing Operation: Each slice operation takes (O(1)) as we’re accessing predefined indices.
• Join Operation: Joining all substrings takes (O(n)), where (n) is the length of the string.

Thus, the total time complexity is O(n).

🚀 Space Complexity

• Auxiliary Space: The result list stores all substrings, requiring (O(n)) space.
• Therefore, the space complexity is O(n).

🔥 Optimized Solution 🚀

The basic solution is already optimal in terms of time and space complexity since it processes the string and indices efficiently. Let’s demonstrate its robustness with a larger-scale example and discuss edge cases.

📚 Example Walkthrough with Higher Numbers

Let’s take a longer string and a more extensive spaces array:

Input:

s = "IHopeYouAreEnjoyingThisChallenge"
spaces = [1, 5, 8, 12, 15, 23, 30]

Execution Steps:

1. Start with index = 0 and result = [].
2. Iterate over each index in spaces:
   • Add the substring s[0:1] → "I", update index = 1.
   • Add the substring s[1:5] → "Hope", update index = 5.
   • Add the substring s[5:8] → "You", update index = 8.
   • Add the substring s[8:12] → "Are", update index = 12.
   • Add the substring s[12:15] → "Enj", update index = 15.
   • Add the substring s[15:23] → "oyingThi", update index = 23.
   • Add the substring s[23:30] → "sChalle", update index = 30.
3. Append the remaining part of the string (s[30:]) → "nge".

Output: "I Hope You Are Enjoying This Challenge"

🧩 Edge Cases

1. Empty Input String:
   • If s = "" and spaces = [], return "".
2. Spaces at Every Character:
   • For s = "abc" and spaces = [1, 2], the output is "a b c".
3. No Spaces to Add:
   • For s = "hello" and spaces = [], the output is "hello".
4. Spaces at the Start or End:
   • For s = "python" and spaces = [0, 6], handle gracefully to avoid slicing errors.

🛠️ Conclusion 🔍

In this blog post, we explored LeetCode #2109 and solved it using a single optimal solution. This problem is a fantastic illustration of string manipulation using slicing, making it highly relevant for technical interviews and string-related tasks.

We hope this walkthrough has been helpful! For more coding insights, keep an eye on our blog. Happy coding! 🌟