How do you solve Split Strings by Separator on LeetCode?

Split Strings by Separator (LeetCode #2788) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using split effectively can simplify string manipulation.

What algorithm or data structure is used to solve Split Strings by Separator?

Split Strings by Separator uses the following concepts: Array, String. The recommended patterns to study are: Array, String.

What are the interview tips for Split Strings by Separator?

Always clarify the requirements before coding. Think about edge cases, such as strings that only contain the separator. Practice writing clean and efficient code.

What are common mistakes when solving Split Strings by Separator?

Not filtering out empty strings, leading to incorrect results. Overcomplicating the solution instead of using built-in functions.

Split Strings by Separator — LeetCode #2788 (Easy)

Tags: Array, String

Related patterns: Array, String

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves iterating through each string in the array and using the split method to break it down by the separator. We then filter out any empty strings from the resulting list.

The time complexity is O(n²) because for each string of length n, we are performing a split operation that can take up to n time, leading to a quadratic relationship in the worst case.

Step 1: Initialize an empty list to store the results.
Step 2: Loop through each string in the input array.
Step 3: For each string, use the split method with the separator to create a list of substrings.
Step 4: Filter out any empty strings from the list of substrings.
Step 5: Add the non-empty substrings to the results list.
Step 6: Return the results list.

1. Input: words = ["one.two.three", "four.five", "six"], separator = "." 2. Initialize result = [] 3. Process "one.two.three": split into ["one", "two", "three"] → result = ["one", "two", "three"] 4. Process "four.five": split into ["four", "five"] → result = ["one", "two", "three", "four", "five"] 5. Process "six": split into ["six"] → result = ["one", "two", "three", "four", "five", "six"] 6. Return result.

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

The optimal solution also involves iterating through the strings but uses a more efficient way to handle the splits and filtering in a single pass, using list comprehension or similar constructs to streamline the process.

The time complexity is O(n) because we are processing each character in the strings once, leading to a linear relationship. The space complexity is O(n) due to the storage of the resulting strings.

Step 1: Initialize an empty list to store the results.
Step 2: Loop through each string in the input array.
Step 3: Use the split method to break the string into parts based on the separator.
Step 4: Use a list comprehension to filter out empty strings and add the valid parts directly to the results list.
Step 5: Return the results list.

Key Insights

Using split effectively can simplify string manipulation.
Filtering out empty strings is crucial to meet the problem requirements.

Common Mistakes

Not filtering out empty strings, leading to incorrect results.
Overcomplicating the solution instead of using built-in functions.

Interview Tips

Always clarify the requirements before coding.
Think about edge cases, such as strings that only contain the separator.
Practice writing clean and efficient code.

#2788

Split Strings by Separator

Easy

Array↗String↗Array↗String↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store the results.
2Step 2: Loop through each string in the input array.
3Step 3: Use the split method to break the string into parts based on the separator.
4Step 4: Use a list comprehension to filter out empty strings and add the valid parts directly to the results list.
5Step 5: Return the results list.

solution.py4 lines

1# Full working Python code
2
3def split_strings(words, separator):
4    return [s for word in words for s in word.split(separator) if s]

ℹ

Complexity note: The time complexity is O(n) because we are processing each character in the strings once, leading to a linear relationship. The space complexity is O(n) due to the storage of the resulting strings.

1Using split effectively can simplify string manipulation.
2Filtering out empty strings is crucial to meet the problem requirements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.