14. Longest Common Prefix

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.



https://leetcode.com/problems/longest-common-prefix/solution/

Solution 2 之后的还没看
今天睡觉之前要完成， 以谷歌面试的bar 要求自己， 一题可以多解的
要多解，分析时空 复杂度， 然后比较trade off， 代码要求bug free ， medium
的题应该在15 分钟之内写完

这个题的考点是什么，突破点在哪， 套路是哪一个




Solution 1 : 
很像 merge sort 的类型， online 算法
class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs == null || strs.length == 0) return "";
        
        if(strs.length == 1) return strs[0];
        String subRes = commonPrefix(strs[0], strs[1]);
        for(int i = 2; i < strs.length; i++){
            subRes = commonPrefix(subRes, strs[i]);
        }
        return subRes;
    }
    private String commonPrefix(String subRes, String cur){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < Math.min(subRes.length(), cur.length()); i++){
            if(subRes.charAt(i) == cur.charAt(i)){
                sb.append(cur.charAt(i));
            }else{
                break;
            }
        }
        return sb.toString();
    }
}







Solution 2 : 
class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs == null || strs.length == 0) return "";
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < strs[0].length(); i++){ // going thru every index of the first word , i = 0, 1
            char c = strs[0].charAt(i); // c = 'a', 'a'
            boolean flag = true;
            for(String word : strs){ // going thru every word and check the char at that certain index 
                // if there is a word whose length is smaller than the current i , break , return the current stringbuilder , else if , the new char is differwent from the char of the first word, break , 
                // if its the same, keep going thru all char at the same index. if boolean is still true 
                // append it 
                if(word.length() <= i || word.charAt(i) != c) flag = false;
               
            }
            if(flag) sb.append(c); // sb = a 
            if(!flag) break;
        }
        return sb.toString();        
    }
}





follow up： 

Given a set of keys S = [s1, s2, s3, s4, .. sn], find the longest common prefix among a string q and S. This LCP query will be called frequently.


Solution: since p changes all the time, the set S doesn’t change, 
We can use a trie to store all the words in S, so it’s gonna be a prefix tree 
Containing all the words from S. So every time when we want to get the common
Prefix, we don’t have to go thru every words in s, instead, we can just traverse the 
Trie tree , it’s more efficient that way 


The end condition when traversing the trie and the word p is when 
there is no common char between p and the trie tree && trie tree has more than one child