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 <= 2000 <= strs[i].length <= 200strs[i] consists of only lowercase English letters if it is non-empty.
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 × 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).
PythonJavaC++GoTypeScriptRustJavaScriptC#PHPRubyC
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](code-box)
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];
}
}(code-box)
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];
}
};(code-box)
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]
}(code-box)
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 '';
}(code-box)
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()
}
}(code-box)
/**
* @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];
};(code-box)
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];
}
}(code-box)
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;
}
}(code-box)
# @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(code-box)
char* longestCommonPrefix(char** strs, int strsSize) {
for (int i = 0; strs[0][i]; i++) {
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != strs[0][i]) {
strs[0][i] = '\0';
return strs[0];
}
}
}
return strs[0];
}(code-box)
