Javascript - Check if an array contains only specified values

Question

How can I optimize a function that checks if an array contains only specified values not using hardcoded values?

Here is the function

function containOnly(value1, value2, array){
  var result;

  for(i = 0; i < array.length; i++){
    if(array[i] != value1 && array[i] != value2){
      result = 0;
      break;
    } else
      result = 1;
  }

  if(result === 0)
    return false;
  else
    return true;
}

console.log(containOnly(1, 2, [2,1,2]));

This function will return true if an array contains specified values. In this function I use if statement to compare two values but how can I use an array of values instead of variables if I want to use more than two values? For example:

console.log(containOnly([1, 2, 3], [2,1,2,3,5]));

Answer 1

You can achieve your requirement using every method by passing an arrow function as argument.

The every() method tests whether all elements in the array pass the test implemented by the provided function.

function containsOnly(array1, array2){
  return array2.every(elem => array1.includes(elem))
}
console.log(containsOnly([1, 2, 3], [2,1,2,3,5]));
console.log(containsOnly([1, 2], [2,1,2,1,1]));

Another solution is to use some method.

function containsOnly(array1, array2){
  return !array2.some(elem => !array1.includes(elem))
}
console.log(containsOnly([1, 2, 3], [2,1,2,3,5]));
console.log(containsOnly([1, 2], [2,1,2,1,1]));

Answer 2

You can simply && up .includes() method.

var arr     = [0,1,3,5,7,9,2,6,8,11,32,53,22,37,91,2,42],
    values1 = [0,2,37,42],
    values2 = [91,99,9],
    checker = ([v,...vs], a) => v !== void 0 ? a.includes(v) && checker(vs, a)
                                             : true;
console.log(checker(values1,arr));
console.log(checker(values2,arr));

This is also efficient than .reduce() since it will stop recursing once the first false value is obtained.