Find array element that is true for ALL elements in the second array-CodePudding

I'm trying to find the smallest common multiple from one array by comparing it to the values in the other by using % num === 0. The answer should be 6, but because each number in the first array is true at least once, they all get returned. How do I find the value that is only true and never false?

let arr = [1, 2, 3]

function test(arr) {
    let x = [1, 2, 3, 4, 5, 6, 7, 8]
    for (let n of arr) {
        return x.filter(k => k % n === 0)
    }
}

console.log(test(arr))

CodePudding user response：

You need loop over the x array and return the first element that gets divided by every value in arr.

let arr = [1, 2, 3];

function test(arr) {
  let x = [1, 2, 3, 4, 5, 6, 7, 8];
  for (let n of x) {
    if (arr.every((a) => n % a === 0)) {
      return n;
    }
  }
}

console.log(test(arr));

You can also simply the solution using Array.prototype.find.

const 
  arr = [1, 2, 3],
  test = (arr) =>
    [1, 2, 3, 4, 5, 6, 7, 8].find((n) => arr.every((a) => n % a === 0));

console.log(test(arr));

Note: If x is not sorted, then you will have to sort it first.

const arr = [1, 2, 3],
  test = (arr) =>
    [8, 7, 6, 5, 4, 3, 2, 1]
      .sort((a, b) => a - b)
      .find((n) => arr.every((a) => n % a === 0));

console.log(test(arr));

Update based on OP's comment

You can use Array.prototype.filter and Array.prototype.some.

const arr = [1, 2, 3],
  test = (arr) =>
    [1, 2, 3, 4, 5, 6, 7, 8].filter((n) => arr.some((a) => n / a === 2));

console.log(test(arr));

CodePudding user response：

If x is sorted can use find and every

let arr = [1, 2, 3]
let x = [1, 2, 3, 4, 5, 6, 7, 8, 12]

let res = x.find(n => arr.every(a => n%a === 0))
console.log(res)

if unsorted x

let arr = [1, 2, 3]
let x = [1, 12, 6, 4, 2, 7, 8]

let res = [...x].sort((a,b)=> a-b).find(n => arr.every(a => n%a === 0))

console.log(res)

CodePudding user response：

Filter and intersect

let arr = [1, 2, 3]
let x = [1, 2, 3, 4, 5, 6, 7, 8]

function test(arr) {
  let common = []
  arr.forEach(n => common.push(x.filter(k => k % n === 0)))
  return common.reduce((acc, cur) => acc.filter(e => cur.includes(e)));
}


console.log(test(arr))

CodePudding user response：

Late to the party answer

You could solved this with one line of code using Array.reduce. Here we initialize it with an empty array and then use a ternary operator to append values that pass the remainder test. Note that we don't need to check for zero (c % n === 0) because we treat the result as a boolean.

x.reduce((p,c) => c % n ? p : p.concat([c]), [])

// TEST

const x = [1, 2, 3, 4, 5, 6, 7, 8];

[1, 2, 3].forEach(n => {

  console.log( 
     "n ="   n,
     x.reduce((p,c) => c % n ? p : p.concat([c]), [])
  );

});