Solving a Leetcode problem daily — Day 6 | Longest Common Prefix
Subhradeep Saha
Posted on May 6, 2024
Solving a Leetcode problem daily — Day 6 | Longest Common Prefix | by Subhradeep Saha | May, 2024 | Medium
Subhradeep Saha ・ ・
Medium
Conquering LeetCode: Finding the Longest Common Prefix
This blog post tackles a classic LeetCode challenge: given an array of strings, write a function that finds the longest string which is a prefix (beginning) of all the strings in the array. We’ll delve into the problem statement, code implementation, step-by-step explanation, real-world applications, and conclude with some key takeaways.
Here is the Leetcode problem link —
https://leetcode.com/explore/learn/card/array-and-string/203/introduction-to-string/1162/
Problem Statement
LeetCode presents a challenge where you’re given an array of strings (strs
). The task is to write a function that returns the longest common prefix shared by all strings in the array.
Example:
Input: strs = ["flower", "flow", "flight"]
Output: "fl"
Explanation:: Here, "fl" is the longest prefix that appears at the beginning
of all three strings.
Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
Constraints:
The number of strings (
strs.length
) can range from 1 to 200.The length of each individual string (
strs[i].length
) can range from 0 to 200.All characters in the strings are lowercase English letters.
Code Implementation
Here’s the C++ code that solves this challenge:
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if (strs.size() == 1) {
return strs[0];
}
string pref = "";
int currIdx = 0;
while (1) {
char ch;
for (int i = 0; i < strs.size(); i++) {
string s = strs[i];
if (currIdx >= s.length()) {
return pref;
}
if (i == 0) {
ch = s[currIdx];
}
if (s[currIdx] != ch) {
return pref;
}
}
pref += strs[0][currIdx++];
}
return pref;
}
};
Breakdown of the Code
This code defines a class Solution
with a function longestCommonPrefix
that takes the array of strings (strs
) as input and returns the longest common prefix.
Edge Case: The code first checks if there’s only one string (
strs.size() == 1
) in the input array. In that case, the single string itself is the longest common prefix, so it's returning that(return strs[0]
).Initialization:
pref
(empty string) stores the growing common prefix.currIdx
(integer) keeps track of the current index (position) within the strings being compared.
3. Looping and Comparison:
- An infinite
while(1)
loop continues until the common prefix is found or there's a mismatch.
Inside the loop:
A character variable
ch
is declared to hold the character to be compared across all strings.A
for
loop iterates through each string (strs[i]
) in the array.If
currIdx
reaches the end of the current string (currIdx >= s.length()
), it means there's no more prefix to compare, so the currentpref
is returned.In the first iteration (
i == 0
), the character atcurrIdx
from the first string (strs[0]
) is assigned toch
as the reference point for comparison.If the character at
currIdx
in the current string (s[currIdx]
) doesn't match the reference characterch
, it indicates a mismatch, and the loop terminates, returning the currentpref
.
4. Building the Prefix:
If all characters at currIdx
across all strings match, the character from strs[0]
at currIdx
is appended to the pref
string and simultaneously the currIdx
is also incremented to move to the next character position for the next iteration. (pref += strs[0][currIdx++]
)
5. Return Value:
The loop eventually breaks when a mismatch is found between the strings character at currIdx
or we reached the end of any of the string. The final pref
string containing the longest common prefix is returned.
Lets verify the code with a dry run of a testcase in the next section
Code Dry Run with a Test Case
Let’s use the provided example:
Input: strs = ["flower", "flow", "flight"]
Step 1: Edge Case Check (Skipped)
There are more than one string, so the code proceeds to the main logic.
Step 2: Initialization
pref
= "" (empty string)currIdx
= 0
Step 3: Loop Iteration 1
Character Comparison:
ch
is assigned the first character from the first string (ch = 'f'
)All three strings have ‘f’ at index 0, so the loop continues.
Building the Prefix:
‘f’ is appended to
pref
:pref = "f"
currIdx
is incremented to 1 (currIdx = 1
)
Step 4: Loop Iteration 2
Character Comparison:
- All three strings have ‘l’ at index 1, so the loop continues.
Building the Prefix:
‘l’ is appended to
pref
:pref = "fl"
currIdx
is incremented to 2 (currIdx = 2
)
Step 5: Loop Termination (Mismatch)
Character Comparison:
- The third string (“flight”) has ‘i’ at index 2, while the first two strings have ‘o’. There’s a mismatch.
The loop terminates at this point.
Result:
The final value of pref
is "fl", which is the longest common prefix for all three strings. The function returns "fl"
.
Real-World Applications
Finding the longest common prefix has various applications in real-world scenarios:
File Compression: In data compression techniques like Lempel-Ziv coding, the longest common prefix between repeated strings can be identified and replaced with a reference, reducing file size.
Autocompletion: Search engines and text editors often use the longest common prefix to suggest completions for partially typed words, improving user experience.
Natural Language Processing (NLP): NLP tasks like text classification or stemming (reducing words to their base forms) can utilize common prefixes to group related words or identify word stems.
Conclusion
This blog post explored the LeetCode challenge of finding the longest common prefix using a C++ code. We broke down the problem statement, presented the code with detailed step-by-step explanations, performed a dry run with a test case and discussed real-world applications.
References
File Compression (Lempel-Ziv Coding):https://ethw.org/Milestones:Lempel-Ziv_Data_Compression_Algorithm,_1977 by Wikipedia: This article provides an overview of Lempel-Ziv (LZ) coding, a family of data compression algorithms. It mentions that LZ77, a specific variant, uses the concept of finding repeating phrases and replacing them with references to a dictionary, effectively using longest common prefixes for compression.
Autocompletion:https://searchengineland.com/how-google-autocomplete-works-390257 by Google Research: This blog post discusses various techniques used in search query autocomplete. It mentions that a prefix tree data structure can be employed to efficiently identify prefixes of words stored in the system, enabling suggestions based on the common prefix typed by the user.
Natural Language Processing (NLP):https://en.wikipedia.org/wiki/Stemming by Wikipedia: This article describes stemming algorithms, a technique in NLP used to reduce words to their base forms. Some stemming algorithms, like the Porter Stemmer, rely on identifying common prefixes among words to achieve stemming.
The diagrams in the Dry run section are created using Canva.
If you liked this blog, do applaude by clicking on the clap hands icon and drop down any feedback you have in the comments.
You can follow me https://medium.com/@subhradeep_saha and check out my other posts as well where I explained the solutions of few other interesting Leetcode problems:
Posted on May 6, 2024
Join Our Newsletter. No Spam, Only the good stuff.
Sign up to receive the latest update from our blog.