1464. Maximum Product of Two Elements in an Array

Problem

Given the array of integers nums, you will choose two different indices i and j of that array. Return the maximum value of (nums[i]-1)*(nums[j]-1).   Example 1:

Input: nums = [3,4,5,2]
Output: 12 
Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12. 

Example 2:

Input: nums = [1,5,4,5]
Output: 16
Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.

Example 3:

Input: nums = [3,7]
Output: 12

  Constraints:

• 2 <= nums.length <= 500

• 1 <= nums[i] <= 10^3

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxProduct = function(nums) {
    if (nums.length === 2) {
        return (nums[0] - 1) * (nums[1] - 1);
    }
    var minNegativeNum = 0;
    var secondMinNegativeNum = 0;
    var maxPositiveNum = 0;
    var secondMaxPositiveNum = 0;
    for (var i = 0; i < nums.length; i++) {
        var num = nums[i] - 1;
        if (num < minNegativeNum) {
            secondMinNegativeNum = minNegativeNum;
            minNegativeNum = num;
        } else if (num < secondMinNegativeNum) {
            secondMinNegativeNum = num;
        } else if (num > maxPositiveNum) {
            secondMaxPositiveNum = maxPositiveNum;
            maxPositiveNum = num;
        } else if (num > secondMaxPositiveNum) {
            secondMaxPositiveNum = num;
        }
    }
    return Math.max(
        minNegativeNum * secondMinNegativeNum,
        maxPositiveNum * secondMaxPositiveNum,
    );
};

Explain:

nope.

Complexity:

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