JCSU Unit 4 Problem Set 2 (Click for link to problem statements)
Problem Highlights
- đź’ˇ Difficulty: Easy
- ⏰ Time to complete: 10-15 mins
- 🛠️ Topics: Complexity Analysis, String Search, Python Built-ins
1: U-nderstand
Understand what the interviewer is asking for by analyzing the behavior and efficiency of the function words_with_char()
.
Questions:
- What is the time complexity of
words_with_char()
when analyzing each word in a list for a character?
- What is the space complexity, considering the input, output, and any auxiliary space?
- How does using Python’s
in
operator affect the function’s efficiency compared to manual iteration?
2: M-atch
Match this problem to known complexity analysis concepts and scenarios.
Key Observations:
-
Iterative Search Through Strings:
- Each word is checked for the presence of a character using either manual iteration or the
in
operator.
- Both approaches involve a linear search through each word.
-
Output List Construction:
- Indices of matching words are stored in a new list, which requires additional space proportional to the number of matches.
3: P-lan
Plan the analysis by breaking down the function’s behavior step by step.
Time Complexity:
- Analyze the number of iterations over the list of words.
- Evaluate the cost of searching for the character in each word.
Space Complexity:
- Identify auxiliary space usage, such as the list of matching indices.
- Assess whether any additional data structures are used.
Optimization Discussion:
- Compare the efficiency of Python’s
in
operator versus manual iteration over characters.
4: I-mplement
Implement the analysis with clear justifications.
1. Time Complexity of words_with_char()
-
Overall Complexity: The time complexity of
words_with_char()
is O(n * m).
-
Reasoning:
- The function iterates through each of the (n) words in the list.
- For each word, it performs a linear search for the character
x
. In the worst case, all (m) characters in the word are examined.
- The total complexity is the product of the number of words and the average word length, resulting in (O(n * m)).
2. Space Complexity of words_with_char()
-
Overall Complexity: The space complexity is O(k), where (k) is the number of words containing the character
x
.
-
Reasoning:
- The function uses a list to store the indices of words that match the condition.
- The size of the list depends on the number of matches, which can be at most (n) (if every word matches).
- No other data structures or additional memory are used, so the complexity is proportional to the output size.
3. Optimization Discussion
5: R-eview
Review the scenarios and validate with examples.
-
Input: words = ["apple", "banana", "cherry"], x = "a"
- Expected Output:
[0, 1]
(indices of words containing "a").
- Observed Complexity: (O(n * m)).
-
Input: words = ["xyz", "123", "abc"], x = "z"
- Expected Output:
[0]
(only the first word contains "z").
- Observed Complexity: (O(n * m)).
-
Input: words = ["dog", "cat", "bat"], x = "e"
- Expected Output:
[]
(no matches).
- Observed Complexity: (O(n * m)).
6: E-valuate
Evaluate the performance of words_with_char()
and the trade-offs between manual iteration and the in
operator.
Summary of Complexity:
-
Time Complexity:
- (O(n * m)), where (n) is the number of words and (m) is the average word length.
- Remains the same for both manual iteration and the
in
operator.
-
Space Complexity:
- (O(k)), where (k) is the number of matches.
Trade-offs:
-
Manual Iteration:
- Slightly more control over character processing.
- Can be more verbose and prone to errors.
-
in
Operator:
- Cleaner and often faster due to Python’s internal optimizations.
- Preferred for readability and simplicity.