Simple logic algorithm

Question

I was doing some algorithm challenge. I am taking a number, converting it to a string then to an array and checking each of its element one by one. Code below works on javascript. The goal is that, if all of the numbers are the ones in the if statement, then the function should return true. But if there is 7 or any other value, then the function will return false.

var a = myFunction(3);

if(a === true){
    alert("true");
}
else{
    alert("false");
}
function myFunction(number){
    let array = number.toString().split("");
    
    for(var i = 0; i<array.length; i++){
        if(array[i] !== '0' && array[i] !== '1' && array[i] !== '8' && array[i] !== '5' && array[i] !== '2'){
            return false;
        }
    }
    return true;
}

I am stuck on the logic below.

if(array[i] !== '0' && array[i] !== '1' && array[i] !== '8' && array[i] !== '5' && array[i] !== '2')

Why will it not work right if I change '&&' to '||'? I am confused by the fact, if array[i] = 1, then it should be false because array[i] !== 0 or 8 or other numbers on the if statement.

The goal is that, if all of the numbers are the ones in the if statement, then the function should return true. — user10693041
– user10693041, Commented Feb 19, 2021 at 16:15
consider (a != 1 || a != 2), if a is 1 then the "not equal to 2" part is true and the whole thing is true, even though the a != 1 part is false. — James
– James, Commented Feb 19, 2021 at 17:36

Kunal Mukherjee · Accepted Answer · 2021-02-19 16:25:24Z

1

To take logical OR || in combination with De Morgan's laws:

!(a && b) = !a || !b 
!(a || b) = !a && !b

function check(number) {
    let array = number.toString().split("");
    for (var i = 0; i < array.length; i++) {
      if (!(array[i] === '0' || array[i] === '1' || array[i] === '8' || array[i] === '5' || array[i] === '2')) {
        return false;
      }
    }
    return true;
}

console.log(check(18520));
console.log(check(7));

edited Feb 19, 2021 at 16:25

Kunal Mukherjee

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answered Feb 19, 2021 at 16:23

Nina Scholz

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1 Comment

user10693041 Over a year ago

Well it did not exactly answer me, but it helped me to understand it better. Thanks!

Jared Farrish · Accepted Answer · 2021-02-19 16:34:52Z

0

const inclusive = set => test => (test + '')
    .split('')
    .filter(value => set.includes(parseInt(value, 10)))
    .length > 0

let test = inclusive([0,1,2,5,8])

console.log('3: ' + test(3))
console.log('321: ' + test(321))
console.log('830: ' + test(830))

answered Feb 19, 2021 at 16:34

Jared Farrish

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