Here is the solution:
Using ES7 (ECMAScript 2016):
const result = PlayerTwo.every(val => PlayerOne.includes(val));
Snippet:
const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C'];
const result = PlayerTwo.every(val => PlayerOne.includes(val));
console.log(result);Using ES5 (ECMAScript 2009):
var result = PlayerTwo.every(function(val) {
return PlayerOne.indexOf(val) >= 0;
});
Snippet:
var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];
var result = PlayerTwo.every(function(val) {
return PlayerOne.indexOf(val) >= 0;
});
console.log(result);Here is answer the question at the comment below:
How do we handle duplicates?
Solution: It is enough to add to the above solution, the accurate condition for checking the number of adequate elements in arrays:
const result = PlayerTwo.every(val => PlayerOne.includes(val)
&& PlayerTwo.filter(el => el === val).length
<=
PlayerOne.filter(el => el === val).length
);
Snippet for first case:
const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C'];
const result = PlayerTwo.every(val => PlayerOne.includes(val)
&& PlayerTwo.filter(el => el === val).length
<=
PlayerOne.filter(el => el === val).length
);
console.log(result);Snippet for second case:
const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C', 'C'];
const result = PlayerTwo.every(val => PlayerOne.includes(val)
&& PlayerTwo.filter(el => el === val).length
<=
PlayerOne.filter(el => el === val).length
);
console.log(result);