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###### tags: `Leetcode`
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# Leetcode 1464. Maximum Product of Two Elements in an Array

[link](https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/)

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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

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題意: 給一個array, 選兩個前二大的數字作運算

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Sort the array and then multiply the last two numbers.

#### Solution 1

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        nums.sort()
        return (nums[-1] - 1) * (nums[-2] - 1)
        

O(T): O(nlogn)
O(S): O(1)

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We also can iterate over the array, and store top two maximum numbers.

#### Solution 2

class Solution:
    def XXX(self, nums: List[int]) -> int:
        largestNum = 0
        secondNum = 0
        for i in range(len(nums)):
            if nums[i] > largestNum:
                secondNum = largestNum
                largestNum = nums[i]
            elif nums[i] > secondNum:
                secondNum = nums[i]
            else:
                continue
                
            print(largestNum)
            print(secondNum)

        return (largestNum - 1) * (secondNum - 1)
        

O(T): O(n)
O(S): O(1)