Sometimes O(n^2) is better than O(n)

I have seen evaluators of coding tests blindly believing that
O(1) is faster than O(log(n)) is faster than O(n) is faster than O(n^2) is faster than O(2^n) is faster than O(n!)
and sometimes judging the candidates wrong. I will give here an example where O(n^2) is better than O(n)

PROBLEM

14. Longest Common Prefix

Easy

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 letters a-z.

SOLUTION in O(n): https://repl.it/@VinitKhandelwal/longest-prefix-javascript-in-n-time

const longestPrefix = arr => {
  if (arr.length === 1) {
    return arr[0];
  } else if (arr.length === 0) {
    return "";
  }
  let check = true
  let char;
  let condition = "(arr[0][j] === arr[1][j])";
  for (let i = 2; i < arr.length; i++) {
    condition += ` && (arr[0][j] === arr[${i}][j])`;
  }
  let end = 0;
  let j = 0;
  while (eval(condition) && arr[0][j] !== undefined) {
    end++;
    j++;
  }
  return arr[0].slice(0, end);
}

Test Input

console.log(longestPrefix(["Jabine","Jaabinder","Jabong"]));

Output

Jab

Takes about 150 ms to execute.

SOLUTION in O(n^2): https://repl.it/@VinitKhandelwal/longest-prefix-javascript

const longestPrefix = arr => {
  if (arr.length === 0) {
    return "";
  }
  if (arr.length === 1) {
    return arr[0];
  }
  let end = 0;
  let check = false
  for (let j = 0; j < arr[0].length; j++){
    for (let i = 1; i < arr.length; i++) {
      if (arr[0][j] !== arr[i][j]) {
        check = true;
        break;
      }
    }
    if (check) {
      break;
    }
    end++;
  }
  return (arr[0].slice(0, end))
}

Test Input

console.log(longestPrefix(["Jabine", "Jabinder", "Jabbong"]))

Output

Jab

Takes about 50 ms to execute.