LeetCode Python Solutions: 14. longest common prefix
ZeroStress LeetCode Python Solutions: 14. longest common prefix #python #leetcode
1. Problem Understanding:
Problem Statement: You are given an array of strings, and you need to find the longest common prefix among them.
Edge Case: If there's no common prefix, return an empty string.
2. Looping and Iteration:
The solution uses a for loop to iterate through the characters of the reference string (base).
It also employs a nested for loop to compare characters in each string in the array with the reference string.
3. String Comparison:
The code compares characters at the same index in different strings to find any mismatches.
The logic uses conditional statements to check for the end of a string and character mismatches.
4. Conditional Statements:
The if statements are used to:
Check if the input list strs is empty. If it is, an empty string is returned immediately to handle this edge case.
Determine if a character mismatch has occurred or if we've reached the end of a string.
5. Slicing Strings:
base[:i] is used to extract a substring from the base string. It returns the characters from the beginning of the string up to (but not including) the i-th character.
This is how the longest common prefix is extracted and returned when a mismatch or string end is encountered.
6. Logic of Finding Common Prefix:
The solution compares characters in a step-by-step manner. It stops as soon as a character mismatch is detected or when the end of any string is reached.
The common prefix is built character by character and returned when the loop exits.
7. Edge Case Handling:
The code handles the case when the input list is empty by returning an empty string right away, avoiding any further processing.
8. Time and Space Complexity:
Time Complexity:
The time complexity of this solution is O(N * M), where:
N is the number of strings in the input array strs.
M is the length of the longest string in the array.
Here's why:
We iterate through the characters of the base string, which has a length of M.
In the worst case, we may need to compare each character of the base string with the corresponding character in all the other N-1 strings in strs.
So, for each character in the base string, we have to perform N-1 comparisons.
Therefore, the overall time complexity is O(M * (N-1)), which simplifies to O(N * M) because we drop the constant factor.
Space Complexity:
The space complexity of this solution is O(1), which means it uses a constant amount of additional space that does not depend on the input size.
Here's why:
We have a constant number of variables used for indexing, comparisons, and storing the reference string base.
We don't use any additional data structures like lists or dictionaries that grow with the input size.
Regardless of the size of the input array or the length of the strings, the space used by this algorithm remains constant.
9. Algorithmic Efficiency:
The algorithm minimizes unnecessary character comparisons by stopping as soon as a mismatch is found.
This ensures efficient execution, especially when dealing with long strings or a large number of strings.
10. Python Language Features:
Understanding Python's syntax, string manipulation, and list indexing is essential to grasp the code's mechanics.