programming issue: unable to resolve

The issue at hand involves creating a function that can determine the presence of all the letters from the second element in the array within the first element. For example, when given the arguments ["hello", "hey"], the function should return false because the word "hello" does not include the letter "y."

I attempted to solve this using the following code, however, it appears to be ineffective and I am unsure of the reason:

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

Answer №1

Instead of returning after the first successful check, consider changing the code to return false at the first unsuccessful check:

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

console.log(mutation(['hello', 'hey']));
console.log(mutation(['hello', 'helo']));

Answer №2

Here is a solution that could assist you:

function checkMutation(array){
var flag = true;
array[1].split('').forEach(letter => {
if(!array[0].includes(letter)){
flag = false;
}
})
return flag;
}
checkMutation(['hello','h']);

Answer №3

Here's a possible solution:

function checkMutation(arr) {
   var testString = arr[1].toLowerCase();
   var targetString = arr[0].toLowerCase();
   for (var i = 0; i < testString.length; i++) {
     if (!targetString.includes(testString[i])) {
       return false;
     }      
   }
   return true;
}

Answer №4

A unique and innovative approach to solving algorithms using functional programming in ES6...

const array1 = ['hello', 'ole'];
const array2 = ['hello', 'helloeloe']
const array3 = ['hello', 'hey']


const mutation = (array) => {
  const [target, test] = array.map(
    item => item.toLowerCase().split('')
  )
  return !test.filter(item => !target.includes(item)).length;
}

console.log(mutation(array1));
console.log(mutation(array2));
console.log(mutation(array3));

Answer №5

If you're looking for a modern, easily readable solution to your problem that avoids any cycles, consider the following alternative:

let words = ['hello', 'hey']

function checkStrings (arrayOfWords) {
  let searchIn = [...arrayOfWords[0]]
  let theseLetters = [...arrayOfWords[1]]
  const condition = letter => searchIn.includes(letter)

  return theseLetters.every(condition)
}

console.log(checkStrings(words))

For a more unconventional approach:

let words = ['hello', 'hey']

function compare (a) {
  let [i, t] = a.map(v => [...v])
  return t.every(l => i.includes(l))
}

console.log(compare(words))

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