Bonfire Mutations

Author

Created by Rafase282

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Details

• Difficulty: 1/5

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.

For example, ['hello', 'Hello'], should return true because all of the letters in the second string are present in the first, ignoring case.

The arguments ['hello', 'hey'] should return false because the string 'hello' does not contain a 'y'.

Lastly, ['Alien', 'line'], should return true because all of the letters in 'line' are present in 'Alien'.

Remember to use Read-Search-Ask if you get stuck. Try to pair program. Write your own code.

Useful Links

• Array.indexOf()

Problem Script:

function mutation(arr) {
  return arr;
}

mutation(['hello', 'hey']);

Explanation:

You basically have to check if the string in the first parameter contains all the characters found on the second parameter.

Hint: 1

First make the arguments lowercase and arrays so we can compare without having to worry about the case.

Hint: 2

Use indexOf() to check if the letter of the second word is on the first.

Hint: 3

You will need a way to check if all the words are in the first one, like keeping a counter and checking for a default value.

My code:

function mutation(arr) {

    var first = arr[0].toLowerCase().split('');
    var second = arr[1].toLowerCase().split('');
    var temp = 0;
    // Check every character and if the index is found add one
    for (var s in second){
        if (first.indexOf(second[s]) > -1) {
            temp+= 0;
        } else
            temp+=1;
        }
    if (temp === 0)
        return true;
    else
        return false;
}

My Code Explanation:

• take the first and second parameter and make them lowercase and split them by characters.
• Create a temp variable to keep count.
• For each letter in the second parameter, check if it is found on the first, if so then add 0 to the tamp variable, otherwise add 1.
• Outside the loop, if temp is 0 then return true, otherwise false, this acts as an on and off. If all the letters were added then we just get 0, otherwise there will be at least a 1 which means one letter was not found.

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