Maximum Product of Three Numbers

#algorithms #maximum #product #javascript

Platforms to practice

Implementation

const maxProduct = arr => {
  if (arr.length < 3) return false;
  const max3 = [-Infinity, -Infinity, -Infinity];
  const min2 = [Infinity, Infinity];

  for (let i = 0; i < arr.length; ++i) {
    const n = arr[i];
    if (n > max3[0]) {
      max3[0] = n;
      max3.sort((a, b) => a - b);
    }
    if (n < min2[0]) {
      min2[0] = n;
      min2.sort((a, b) => b - a);
    }
  }
  return max3[2] * Math.max(min2[0] * min2[1], max3[0] * max3[1]);
};

Complexity

Time complexity: O(n)
Space complexity: O(1)

Explanation

Loop through array, while keeping 3 max elements and 2 min elements.
We want 2 min elements because min element can have bigger absolute value. When we calculating product always 2 negative numbers give 1 positive, so we need only 2 min to check if their product is not bigger that 2 smaller numbers of our 3 max.