Discover how a simple trick can save you from endless letter-by-letter guessing!
Why Longest Common Prefix in DSA Javascript?
Imagine you have a list of words like "flower", "flow", "flight" and you want to find the part they all start with. Doing this by checking each letter one by one for every word can be tiring and confusing.
Manually comparing each letter across all words is slow and easy to mess up, especially if the list is long or words are long. You might miss a letter or compare too many times, making it frustrating and error-prone.
The Longest Common Prefix method quickly finds the shared beginning part of all words by smartly comparing letters only as needed. It saves time and avoids mistakes by stopping as soon as a difference is found.
function findPrefix(words) {
let prefix = '';
for (let i = 0; i < words[0].length; i++) {
let char = words[0][i];
for (let word of words) {
if (word[i] !== char) return prefix;
}
prefix += char;
}
return prefix;
}function longestCommonPrefix(words) {
if (!words.length) return '';
let prefix = words[0];
for (let i = 1; i < words.length; i++) {
while (!words[i].startsWith(prefix)) {
prefix = prefix.slice(0, -1);
if (!prefix) return '';
}
}
return prefix;
}This lets you quickly find the shared start of many words, which helps in search engines, autocomplete, and organizing data efficiently.
When you type in a search box, the system suggests words that start with what you typed. Longest Common Prefix helps find those suggestions fast by identifying the common start among many options.
Manually checking letters is slow and error-prone.
Longest Common Prefix finds shared starts efficiently.
Useful for autocomplete, search, and data grouping.
Practice
Solution
Step 1: Understand the problem definition
The Longest Common Prefix means the longest substring at the start of all strings that is the same.Step 2: Compare options with definition
Only The longest starting substring common to all strings matches this definition exactly.Final Answer:
The longest starting substring common to all strings -> Option BQuick Check:
Longest Common Prefix = longest starting substring [OK]
- Confusing prefix with suffix
- Looking for substring anywhere, not just start
- Thinking about unique substrings
longestCommonPrefix that takes an array strs as input?Solution
Step 1: Recall JavaScript function syntax
JavaScript functions start with the keyword 'function' followed by name and parentheses.Step 2: Check each option
function longestCommonPrefix(strs) { } matches JavaScript syntax. Options B and C are from other languages. function: longestCommonPrefix(strs) { } is invalid syntax.Final Answer:
function longestCommonPrefix(strs) { } -> Option CQuick Check:
JS function syntax = function name(params) { } [OK]
- Using Python or other language syntax
- Adding colon after function name
- Using 'func' instead of 'function'
function longestCommonPrefix(strs) {
if (strs.length === 0) return "";
let prefix = strs[0];
for (let i = 1; i < strs.length; i++) {
while (strs[i].indexOf(prefix) !== 0) {
prefix = prefix.substring(0, prefix.length - 1);
if (prefix === "") return "";
}
}
return prefix;
}
console.log(longestCommonPrefix(["flower", "flow", "flight"]));Solution
Step 1: Initialize prefix and loop through strings
Prefix starts as "flower". Check "flow": "flower" is not prefix, shorten to "flowe", then "flow", which matches start of "flow".Step 2: Check next string "flight"
"flow" is not prefix of "flight", shorten to "flo", no match; then "fl", which matches start of "flight".Final Answer:
"fl" -> Option AQuick Check:
Common prefix of flower, flow, flight = "fl" [OK]
- Returning first string as prefix
- Not shrinking prefix correctly
- Confusing substring with suffix
function longestCommonPrefix(strs) {
if (!strs.length) return "";
let prefix = strs[0];
for (let i = 1; i <= strs.length; i++) {
while (strs[i].indexOf(prefix) !== 0) {
prefix = prefix.substring(0, prefix.length - 1);
if (prefix === "") return "";
}
}
return prefix;
}Solution
Step 1: Check for loop bounds
The for loop uses i <= strs.length, which causes out-of-bounds access at last iteration.Step 2: Correct loop condition
Changing to i < strs.length prevents accessing undefined strs[i].Final Answer:
The for loop condition should be i < strs.length, not i <= strs.length -> Option AQuick Check:
Loop index must be less than array length [OK]
- Off-by-one error in loop
- Wrong initialization of prefix
- Misusing string methods
Solution
Step 1: Check for empty array or empty strings
If the array is empty or any string is empty, the longest common prefix must be empty string.Step 2: Return empty string early
Returning "" immediately avoids errors and matches expected output.Final Answer:
Return an empty string immediately if the array is empty or any string is empty -> Option DQuick Check:
Empty input means empty prefix [OK]
- Not handling empty strings causing errors
- Ignoring empty strings and causing wrong prefix
- Throwing errors instead of returning empty string
Ready to prove it?
Pass the test and this pattern moves unknown
