# LC 2708. Maximum Strength of a Group

### [Problem link](https://leetcode.com/problems/maximum-strength-of-a-group/)

###### tags: `leedcode` `python` `medium`

You are given a **0-indexed** integer array <code>nums</code> representing the score of students in an exam. The teacher would like to form one **non-empty** group of students with maximal **strength** , where the strength of a group of students of indices <code>i₀</code>, <code>i₁</code>, <code>i₂</code>, ... , <code>i_k</code> is defined as <code>nums[i₀] * nums[i₁] * nums[i₂] * ... * nums[i_k]</code>.

Return the maximum strength of a group the teacher can create.

**Example 1:**

Input: nums = [3,-1,-5,2,5,-9]
Output: 1350
Explanation: One way to form a group of maximal strength is to group the students at indices [0,2,3,4,5]. Their strength is 3 * (-5) * 2 * 5 * (-9) = 1350, which we can show is optimal.

**Example 2:**

Input: nums = [-4,-5,-4]
Output: 20
Explanation: Group the students at indices [0, 1] . Then, we'll have a resulting strength of 20. We cannot achieve greater strength.

**Constraints:**

- <code>1 <= nums.length <= 13</code>
- <code>-9 <= nums[i] <= 9</code>

## Solution 1

class Solution:
    def maxStrength(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]
        
        neg = []
        pos = []
        for num in nums:
            if num > 0:
                pos.append(num)
            elif num < 0:
                neg.append(num)
        
        if len(neg) % 2:
            neg.remove(max(neg))
            
        if not neg and not pos:
            return 0
        
        res = 1
        for p in pos:
            res *= p
        for n in neg:
            res *= n
            
        return res
        

>### Complexity
>n = nums.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(n) |

## Note
分成positive, negative處理.