Algorithm Mutations

Algorithm Mutations

Problem Explanation:

• Return true if the string in the first element of the array contains all of the letters of the string in the second element of the array..

Hint: 1

• If everything is lowercase it will be easier to compare.

Hint: 2

• Our strings might be easier to work with if they were arrays of characters.

Hint: 3

• A loop might help. Use indexOf() to check if the letter of the second word is on the first.

Spoiler Alert!

Solution ahead!

Code Solutions:

First solution (Procedural):

function mutation(arr) {
  var test = arr[1].toLowerCase();
  var target = arr[0].toLowerCase();
  for (i=0;i<test.length;i++) {
    if (target.indexOf(test[i]) < 0)
      return false;
  }
  return true;
 }

🚀 Run Code

Second Solution (Declarative):

function mutation(arr) {
 return arr[1].toLowerCase()
              .split('')
              .every(function (letter){
                return arr[0].toLowerCase()
                             .indexOf(letter) != -1;
              });
}

🚀 Run Code

Code Explanations:

First solution (Procedural):

First we make the two strings in the array lowercase. test will hold what we are looking for in target.
Then we loop through our test characters and if any of them is not found we return false.

If they are all found, the loop will finish without returning anything and we get to return true.

Second solution (Declarative):

Grab the second string, lowercase and turn it into an array; then make sure every one of its letters is a part of the lowercased first string.

Every will basically give you letter by letter to compare, which we do by using indexOf on the first string. indexOf will give you -1 if the current letter is missing. We check that not to be the case, for if this happens even once every will be false.

Credits:

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