Watch 8 video solutions for Maximum Strength of a Group, a medium level problem involving Array, Dynamic Programming, Backtracking. This walkthrough by Tech Courses has 728 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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] <= 9Problem Overview: You are given an integer array where each value represents the strength of a member. The task is to select a non-empty subset whose product is maximized. Because the array can contain positive numbers, negative numbers, and zeros, choosing which elements to multiply becomes a careful balancing act.
Approach 1: Backtracking (O(2^n) time, O(n) space)
This method enumerates every possible non-empty subset and computes the product for each. You recursively decide whether to include or exclude each element while maintaining the current product. The maximum product seen during enumeration becomes the result. This approach is conceptually simple and demonstrates the full search space, but it becomes impractical as the array grows because the number of subsets doubles with every new element.
Backtracking is useful when explaining the brute-force baseline during interviews. It also helps illustrate why smarter strategies are needed for larger inputs. The recursion depth is at most n, giving O(n) auxiliary space. However, the exponential O(2^n) runtime makes it unsuitable for production scenarios.
Approach 2: Sorting and Greedy Multiplication (O(n log n) time, O(1) space)
The optimal strategy relies on how multiplication behaves with positive and negative numbers. Positive numbers always increase the product, while negative numbers can increase it only when paired together. Start by sorting the array so negatives and positives are grouped. Multiply all positive numbers. For negatives, pair them from the smallest values (largest magnitude) because the product of two negatives becomes positive.
If there is an odd count of negatives, skip the one closest to zero because including it would flip the sign of the product. Zeros act as neutralizers: if every other choice leads to a negative product, selecting zero can yield a better result. This greedy reasoning ensures you keep only values that contribute positively to the final product.
This approach leverages properties of multiplication rather than exploring every subset. Sorting simplifies pairing logic and ensures the strongest negatives are combined first. The algorithm runs in O(n log n) time due to sorting and uses constant extra space. It commonly appears in problems involving greedy selection and sign management within array processing. Some solutions also connect this reasoning to state transitions similar to dynamic programming maximum-product problems.
Recommended for interviews: Interviewers expect the greedy reasoning. Starting with the backtracking solution shows you understand the brute-force search space, but transitioning to the sorting-based greedy method demonstrates algorithmic insight and the ability to exploit mathematical properties of multiplication.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Backtracking Enumeration | O(2^n) | O(n) | Understanding the brute-force search space or when n is very small |
| Sorting + Greedy Multiplication | O(n log n) | O(1) | General case; efficiently handles positives, negatives, and zeros |