# 2708. Maximum Strength of a Group

## Description

You are given a 0-indexed integer array nums 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 i0, i1, i2, ... , ik is defined as nums[i0] * nums[i1] * nums[i2] * ... * nums[ik​].

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:

• 1 <= nums.length <= 13
• -9 <= nums[i] <= 9

## Solutions

• class Solution {
public long maxStrength(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
if (n == 1) {
return nums[0];
}
if (nums[1] == 0 && nums[n - 1] == 0) {
return 0;
}
long ans = 1;
int i = 0;
while (i < n) {
if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
ans *= nums[i] * nums[i + 1];
i += 2;
} else if (nums[i] <= 0) {
i += 1;
} else {
ans *= nums[i];
i += 1;
}
}
return ans;
}
}

• class Solution {
public:
long long maxStrength(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
if (n == 1) {
return nums[0];
}
if (nums[1] == 0 && nums[n - 1] == 0) {
return 0;
}
long long ans = 1;
int i = 0;
while (i < n) {
if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
ans *= nums[i] * nums[i + 1];
i += 2;
} else if (nums[i] <= 0) {
i += 1;
} else {
ans *= nums[i];
i += 1;
}
}
return ans;
}
};

• class Solution:
def maxStrength(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
if n == 1:
return nums[0]
if nums[1] == nums[-1] == 0:
return 0
ans, i = 1, 0
while i < n:
if nums[i] < 0 and i + 1 < n and nums[i + 1] < 0:
ans *= nums[i] * nums[i + 1]
i += 2
elif nums[i] <= 0:
i += 1
else:
ans *= nums[i]
i += 1
return ans


• func maxStrength(nums []int) int64 {
sort.Ints(nums)
n := len(nums)
if n == 1 {
return int64(nums[0])
}
if nums[1] == 0 && nums[n-1] == 0 {
return 0
}
ans := int64(1)
for i := 0; i < n; i++ {
if nums[i] < 0 && i+1 < n && nums[i+1] < 0 {
ans *= int64(nums[i] * nums[i+1])
i++
} else if nums[i] > 0 {
ans *= int64(nums[i])
}
}
return ans
}

• function maxStrength(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
if (n === 1) {
return nums[0];
}
if (nums[1] === 0 && nums[n - 1] === 0) {
return 0;
}
let ans = 1;
for (let i = 0; i < n; ++i) {
if (nums[i] < 0 && i + 1 < n && nums[i + 1] < 0) {
ans *= nums[i] * nums[i + 1];
++i;
} else if (nums[i] > 0) {
ans *= nums[i];
}
}
return ans;
}