What if you could instantly find the shared start of many words without checking each one painfully?
Why Longest Common Prefix in DSA Java?
Imagine you have a list of words and you want to find the first few letters they all share. Doing this by checking each letter one by one for every word can be tiring and confusing.
Checking each word manually is slow and easy to mess up. You might forget to compare all words or miss a letter, making your answer wrong or incomplete.
The Longest Common Prefix method quickly finds the shared starting letters among all words by comparing them smartly, saving time and avoiding mistakes.
String prefix = words[0]; for (String word : words) { while (!word.startsWith(prefix)) { prefix = prefix.substring(0, prefix.length() - 1); } }
public String longestCommonPrefix(String[] words) {
if (words == null || words.length == 0) return "";
for (int i = 0; i < words[0].length(); i++) {
char c = words[0].charAt(i);
for (int j = 1; j < words.length; j++) {
if (i == words[j].length() || words[j].charAt(i) != c) {
return words[0].substring(0, i);
}
}
}
return words[0];
}This lets you quickly find the shared start of many words, helping in tasks like auto-complete, search suggestions, and DNA sequence analysis.
When you type in a search box, the system suggests words that start with the letters you typed. Finding the longest common prefix helps the system know what letters all suggestions share.
Manual checking is slow and error-prone.
Longest Common Prefix finds shared starts efficiently.
Useful for search, typing aids, and pattern matching.
Practice
Solution
Step 1: Understand the prefix concept
A prefix is a substring starting at the beginning of a string.Step 2: Define longest common prefix
The longest common prefix is the longest substring starting at the beginning that all strings share.Final Answer:
The longest starting substring common to all strings -> Option CQuick Check:
Prefix means start of string = longest common start [OK]
- Confusing prefix with substring anywhere
- Thinking it is the longest suffix
- Mixing prefix with shortest substring
strs?Solution
Step 1: Identify return type for prefix
The longest common prefix is a string, so return type must be String.Step 2: Check parameter type
The input is an array of strings, so parameter type is String[].Final Answer:
public String longestCommonPrefix(String[] strs) -> Option AQuick Check:
Return String, input String[] = public String longestCommonPrefix(String[] strs) [OK]
- Using void return type instead of String
- Returning int instead of String
- Using List<String> instead of String[]
public class Test {
public static String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) return "";
String prefix = strs[0];
for (int i = 1; i < strs.length; i++) {
while (!strs[i].startsWith(prefix)) {
prefix = prefix.substring(0, prefix.length() - 1);
if (prefix.isEmpty()) return "";
}
}
return prefix;
}
public static void main(String[] args) {
String[] input = {"flower", "flow", "flight"};
System.out.println(longestCommonPrefix(input));
}
}Solution
Step 1: Initialize prefix and check each string
Start with prefix = "flower". Check "flow": "flow" starts with "flower"? No, shorten prefix to "flowe", then "flow", which matches.Step 2: Check next string "flight"
"flight" starts with "flow"? No, shorten prefix to "flo", then "fl", which matches.Final Answer:
"fl" -> Option BQuick Check:
Common start of all strings = "fl" [OK]
- Returning first string without checking others
- Not shortening prefix correctly
- Returning empty string too early
public String longestCommonPrefix(String[] strs) {
if (strs.length == 0) return null;
String prefix = strs[0];
for (int i = 1; i <= strs.length; i++) {
while (!strs[i].startsWith(prefix)) {
prefix = prefix.substring(0, prefix.length() - 1);
if (prefix.isEmpty()) return "";
}
}
return prefix;
}Solution
Step 1: Check loop bounds
The for loop uses i <= strs.length, which causes ArrayIndexOutOfBoundsException because valid indices are 0 to strs.length - 1.Step 2: Correct loop condition
Change to i < strs.length to avoid accessing invalid index.Final Answer:
The loop condition should be i < strs.length, not i <= strs.length -> Option DQuick Check:
Array indices go from 0 to length-1 [OK]
- Using i <= length causing runtime error
- Returning null instead of empty string
- Incorrect substring usage
String[] words = {"interview", "interrupt", "integrate", "integral"};, which of the following modifications to the longest common prefix algorithm will correctly find the prefix "int" even if some strings are empty or null?Solution
Step 1: Understand problem with null or empty strings
Null or empty strings cause errors or incorrect prefix results if not handled.Step 2: Modify algorithm to skip or handle null/empty
Check each string before comparing; skip or handle null/empty to avoid errors and find correct prefix.Final Answer:
Add checks to skip null or empty strings before comparing prefixes -> Option AQuick Check:
Handle null/empty strings safely to find correct prefix [OK]
- Not handling null causing exceptions
- Returning first string without checks
- Starting prefix from wrong string
Ready to prove it?
Pass the test and this pattern moves unknown
