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
Solution 1: Binary Enumeration
The problem is actually to find the maximum product of all subsets. Since the length of the array does not exceed $13$, we can consider using the method of binary enumeration.
We enumerate all subsets in the range of $[1, 2^n)$, and for each subset, we calculate its product, and finally return the maximum value.
The time complexity is $O(2^n \times n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
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Solution 2: Sorting + Greedy
First, we can sort the array. Based on the characteristics of the array, we can draw the following conclusions:
- If there is only one element in the array, then the maximum strength value is this element.
- If there are two or more elements in the array, and $nums[1] = nums[n - 1] = 0$, then the maximum strength value is $0$.
- Otherwise, we traverse the array from small to large. If the current element is less than $0$ and the next element is also less than $0$, then we multiply these two elements and accumulate the product into the answer. Otherwise, if the current element is less than or equal to $0$, we skip it directly. If the current element is greater than $0$, we multiply this element into the answer. Finally, we return the answer.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array.
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