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.
My github reference
https://github.com/gkucmierz/algorithms/blob/master/problems/maximum_product_of_three_numbers.js
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