# 14. Longest Common Prefix

## Description

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input: strs = ["flower","flow","flight"]
Output: "fl"


Example 2:

Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.


Constraints:

• 1 <= strs.length <= 200
• 0 <= strs[i].length <= 200
• strs[i] consists of only lowercase English letters.

## Solutions

Solution 1: Character Comparison

We use the first string $strs[0]$ as a benchmark, and compare whether the $i$-th character of the subsequent strings is the same as the $i$-th character of $strs[0]$. If they are the same, we continue to compare the next character. Otherwise, we return the first $i$ characters of $strs[0]$.

If the traversal ends, it means that the first $i$ characters of all strings are the same, and we return $strs[0]$.

The time complexity is $O(n \times m)$, where $n$ and $m$ are the length of the string array and the minimum length of the strings, respectively. The space complexity is $O(1)$.

• class Solution {
public String longestCommonPrefix(String[] strs) {
int n = strs.length;
for (int i = 0; i < strs[0].length(); ++i) {
for (int j = 1; j < n; ++j) {
if (strs[j].length() <= i || strs[j].charAt(i) != strs[0].charAt(i)) {
return strs[0].substring(0, i);
}
}
}
return strs[0];
}
}

• class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
int n = strs.size();
for (int i = 0; i < strs[0].size(); ++i) {
for (int j = 1; j < n; ++j) {
if (strs[j].size() <= i || strs[j][i] != strs[0][i]) {
return strs[0].substr(0, i);
}
}
}
return strs[0];
}
};

• class Solution:
def longestCommonPrefix(self, strs: List[str]) -> str:
for i in range(len(strs[0])):
for s in strs[1:]:
if len(s) <= i or s[i] != strs[0][i]:
return s[:i]
return strs[0]


• func longestCommonPrefix(strs []string) string {
n := len(strs)
for i := range strs[0] {
for j := 1; j < n; j++ {
if len(strs[j]) <= i || strs[j][i] != strs[0][i] {
return strs[0][:i]
}
}
}
return strs[0]
}

• function longestCommonPrefix(strs: string[]): string {
const len = strs.reduce((r, s) => Math.min(r, s.length), Infinity);
for (let i = len; i > 0; i--) {
const target = strs[0].slice(0, i);
if (strs.every(s => s.slice(0, i) === target)) {
return target;
}
}
return '';
}


• /**
* @param {string[]} strs
* @return {string}
*/
var longestCommonPrefix = function (strs) {
for (let j = 0; j < strs[0].length; j++) {
for (let i = 0; i < strs.length; i++) {
if (strs[0][j] !== strs[i][j]) {
return strs[0].substring(0, j);
}
}
}
return strs[0];
};


• class Solution {
/**
* @param String[] $strs * @return String */ function longestCommonPrefix($strs) {
$rs = ''; for ($i = 0; $i < strlen($strs[0]); $i++) { for ($j = 1; $j < count($strs); $j++) { if ($strs[0][$i] !=$strs[$j][$i]) {
return $rs; } }$rs = $rs .$strs[0][$i]; } return$rs;
}
}

• # @param {String[]} strs
# @return {String}
def longest_common_prefix(strs)
return '' if strs.nil? || strs.length.zero?

return strs[0] if strs.length == 1

idx = 0
while idx < strs[0].length
cur_char = strs[0][idx]

str_idx = 1
while str_idx < strs.length
return idx > 0 ? strs[0][0..idx-1] : '' if strs[str_idx].length <= idx

return '' if strs[str_idx][idx] != cur_char && idx.zero?
return strs[0][0..idx - 1] if strs[str_idx][idx] != cur_char
str_idx += 1
end

idx += 1
end

idx > 0 ? strs[0][0..idx] : ''
end


• public class Solution {
public string LongestCommonPrefix(string[] strs) {
int n = strs.Length;
for (int i = 0; i < strs[0].Length; ++i) {
for (int j = 1; j < n; ++j) {
if (i >= strs[j].Length || strs[j][i] != strs[0][i]) {
return strs[0].Substring(0, i);
}
}
}
return strs[0];
}
}

• impl Solution {
pub fn longest_common_prefix(strs: Vec<String>) -> String {
let mut len = strs
.iter()
.map(|s| s.len())
.min()
.unwrap();
for i in (1..=len).rev() {
let mut is_equal = true;
let target = strs[0][0..i].to_string();
if strs.iter().all(|s| target == s[0..i]) {
return target;
}
}
String::new()
}
}