Get all number pairs which [[a, b], a * b = value], for a given value

allMultiplierPairsOfArr2.js

/**
 * mostly ES5 without featured code
 * except for let and arrow functions which are better for reading and understanding
 *
 * FYI
 * 1) a => a > 0 equals function (a) { return a > 0; } and is just a syntax sugar
 * 2) let is used as an example of best practices
 * 3) map is used to change the element to the result
 *    [-3, 1, 15].map(a => a < 0 ? [a, 0] : [0, a]) => [[-3, 0], [0, 1], [0, 15]]
 * 4) filter is used to get only those elements which function is true for the condition
 *    [-3, 0, 15].filter(a => a > 0) => [15]
 * 5) reverse is used to change the elements order to the opposite
 *    [1, 2, 3].reverse() => [3, 2, 1]
 * 6) concat pushes the elements of both arrays to the resulted array
 *    [1, 2, 3].concat([4, 5, 6]) => [1, 2, 3, 4, 5, 6]
 *
 */
function findPairsInSorted(arr, value) {
  let results = [];
  let len = arr ? arr.length : 0;
  if (len < 1) {
    return results;
  }
  for (let i = 0, j = len - 1; i < j;) {
    if (value % arr[i] !== 0) {
      i++;
    }
    else if (value / arr[i] < arr[j]) {
      j--;
    }
    else if (value / arr[i] > arr[j]) {
      i++;
    }
    else {
      results.push([arr[i], arr[j]]);
      i++;
      j--;
    }
  }
  return results;
}
 
function solve(arr, value) {
  let sortedArr = arr.sort((a, b) => a - b);
  if (value > 0) {
    let positive = findPairsInSorted(sortedArr.filter(a => a > 0), value);
    let negative = findPairsInSorted(sortedArr.filter(a => a < 0), value);
    return negative.concat(positive);
  }
  if (value < 0) {
    let all = sortedArr.filter(a => a < 0).reverse().concat(sortedArr.filter(a => a > 0));
    return findPairsInSorted(all, value);
  }
  return sortedArr.filter(a => a !== 0).map(a => a < 0 ? [a, 0] : [0, a]);
}

solve([ 3, 4, 5, -2, 10, 11, 12, -1, 0, 7, 8 ], 10);
solve([ 10, 1, -1, -10, 0 ], -10);
solve([ 1, 18, 3, 6, 9, 2 ], 18);
solve([ 10, 1, -1, -10, 0 ], 0);