Longest Common Prefix
(LeetCode Problems)
Problem:
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:
SOLUTION:
/**
* @param {string[]} strs
* @return {string}
*/
var longestCommonPrefix = function(strs) {
if(strs.length == 1) {
return strs[0]
}
let shortName = strs.sort((a,b) => a.length-b.length)[0];
let common = "";
for(let i=0;i<shortName.length;i++) {
let hasLetter = false;
for(let j=0;j<strs.length;j++){
if(shortName[i] != strs[j][i]) {
return common;
} else {
hasLetter = true
}
}
if(hasLetter) {
common += shortName[i];
}
}
return common ? common : "";
}; |
Explanation:
strs is an array and is input value.
Step 1:
if(strs.length == 1) {
return strs[0]
}
/*
* if strs has only one input then that value will result as output.
*/ |
Step 2:
/*
*Longest common prefix will also be the shortest length string in an array. So, let's find out the *shortest item first.
*/
let shortName = strs.sort((a,b) => a.length-b.length)[0]; |
Step 3:
/*
*Now create a variable common as an empty string.
*/
let common = "";
/* Now check every string starting from position 0, if found in every item concat string and continue else exit from the loop.
*/
for(let i=0;i<shortName.length;i++) {
let hasLetter = false;
for(let j=0;j<strs.length;j++){
if(shortName[i] != strs[j][i]) {
return common;
} else {
hasLetter = true
}
}
if(hasLetter) {
common += shortName[i];
}
}
// finally return the common string |
Here is the link to the original question.
