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## 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: ["flower","flow","flight"]
Output: "fl"
```

Example 2:

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

Note:

All given inputs are in lowercase lettersĀ `a-z`.

## Submissions

### Horizontal Scan

One way is to set the prefix to the first string in the array and iteratively shrink it.

```class Solution {
public String longestCommonPrefix(String[] strs) {
if(strs.length == 0) {
return "";
}

String prefix = new String(strs[0]);
for(String str : strs) {
int i = 0;
for(; i < str.length() && i < prefix.length(); i++) {
if(str.charAt(i) != prefix.charAt(i)) {
break;
}
}
prefix = prefix.substring(0, i);
}

return prefix;
}
}```

class Solution {
public String longestCommonPrefix(String[] strs) {
if(strs.length == 0) {
return “”;
}

```    String prefix = new String(strs[0]);
for(String str : strs) {
int i = 0;
for(; i < str.length() && i < prefix.length(); i++) {
if(str.charAt(i) != prefix.charAt(i)) {
break;
}
}
prefix = prefix.substring(0, i);
}

return prefix;
}```

## Vertical Scan

The time complexity depends on the time of scans, obviously, the vertical scan will reduce the time since it can scan the least of the letters.

```class Solution {
public String longestCommonPrefix(String[] strs) {
if(strs.length == 0) return "";
for(int i = 0; ; i++) {
if(strs[0].length() == i) {
return strs[0];
}
char c = strs[0].charAt(i);
for(String str : strs) {
if(str.length() == i) {
return str;
}

if(str.charAt(i) != c) {
return str.substring(0, i);
}
}
}
}
}```