Related to : A function that compare if there is duplicate numbers in a array

bcrypt compare returning false when password contains numbers 
In your beforeValidate function you are passing in 10 as the salt you
need to generate your salt, change that function to this
beforeValidate: function(user, model, cb) {
var salt = bcrypt.genSalt(10);
bcrypt.hash(user.password, salt, function(err, hash) {
if ( err ) { throw err; }
user.password = hash;
cb(null, user);
});
}

Compare two arrays to get the missing array and new array in java 
You can use guava libraries for getting the difference between the
sets.
What you can do is convert bothe the arrays to Set
new HashSet(Arrays.asList(array));
And similalarly for the sencond array as well
Then pass the Sets to guava difference method
public static Sets.SetView difference(Set set1, Set set2)
.If you don't want to use any 3rd party libraries, You can check out
this question
ques

compare elements positions in first array with another array 
Its just a loop.
function checkArrays( arrA, arrB ){
for(var i = 0; i < arrA.length; i++) { if(arrA[i] !== arrB [i])
return arrB [i] };
}
http://jsfiddle.net/hL257bh6/
If you're allowed to use ECMAScript 6
a.find(function(v, i){ return v !== d[i] })

compare Java Double.Max_VALUE with large double numbers returns true 
I think the problem is in this line

v
new Double(Double.MAX_VALUE).equals(new Double(Double.MAX_VALUE 
99999999999999999999999D))
DOUBLE MAX_VALUE = Double.MAX_VALUE  99999999999999999999999D
AND you're comparing it with
new Double(Double.MAX_VALUE  99999999999999999999999D

Find triplets in an array such that sum of two numbers is also a number in the given array 
At first glance, I can't see an algorithm that is less than O(n²).
(Unless it is an extraordinarily clever one)
This problem is very similar to the infamous 3SUM question: Basically,
the idea is that given a set of n elements, are there any triples that
sum to zero?
An algorithm solving 3SUM faster than O(n²) is not known  This is an
open problem in computer science... (See here: http://en.
