Problem Description

You are given a 0-indexed integer array nums representing the scores of students in an exam. The teacher wants to form one non-empty group of students to maximize the group's strength.

The strength of a group is calculated by multiplying all the scores of the selected students together. For example, if you select students at indices i₀, i₁, i₂, ..., iₖ, the strength would be nums[i₀] × nums[i₁] × nums[i₂] × ... × nums[iₖ].

Your task is to find and return the maximum possible strength value that can be achieved by selecting any non-empty subset of students.

Key Points:

• You must select at least one student (non-empty group)
• The strength is the product of all selected students' scores
• The array can contain positive, negative, or zero values
• You need to find the maximum possible product among all possible selections

For example, if nums = [3, -1, -5, 2], you could:

• Select just [3] for strength = 3
• Select [3, 2] for strength = 6
• Select [-1, -5] for strength = 5
• Select [-1, -5, 2] for strength = 10
• Select [3, -1, -5, 2] for strength = 30

The maximum strength would be 30.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem involves an array of student scores, not nodes and edges as in graph problems.

Need to solve for kth smallest/largest?

• No: We're looking for the maximum strength value, not the kth smallest or largest element.

Involves Linked Lists?

• No: The problem uses an array structure, not linked lists.

Does the problem have small constraints?

• Yes: The problem states that the array length does not exceed 13, which is a small constraint. With at most 13 elements, we have at most 2^13 = 8192 possible subsets to consider.

Brute force / Backtracking?

• Yes: Given the small constraints, we can afford to explore all possible non-empty subsets of students to find the maximum product. This is exactly what the solution does - it enumerates all possible combinations.

Conclusion: The flowchart correctly leads us to use a Brute Force / Backtracking approach. The small constraint (n ≤ 13) makes it feasible to enumerate all 2^n - 1 non-empty subsets and calculate their products to find the maximum strength. The solution uses binary enumeration (bit manipulation) to generate all possible subsets, which is an efficient implementation of the brute force approach for this problem size.

Intuition

When we need to find the maximum product from selecting any subset of students, our first thought might be to use a greedy approach - perhaps always picking positive numbers or the largest values. However, this problem has a twist: negative numbers can actually help us! Two negative numbers multiply to give a positive result, so sometimes including negative values can increase our product.

Since we need to consider all possible combinations of students to find the true maximum, and we can't easily predict which combination will give us the best result (due to the interplay between positive and negative numbers), we need to check all possibilities.

The key insight is recognizing that with only 13 elements maximum, we have a manageable number of subsets to check. Each element can either be included or excluded from our group, giving us 2^n total subsets. Since we need at least one student (non-empty group), we have 2^n - 1 valid subsets to evaluate.

Binary enumeration becomes a natural fit here. We can represent each subset as a binary number where each bit position corresponds to whether we include that student or not. For example, with 3 students, the binary number 101 would mean we include the first and third students but not the second.

By iterating through all numbers from 1 to 2^n - 1, we're essentially generating all possible non-empty subsets. For each subset (represented by a number i), we check each bit position to see which students to include, calculate the product of their scores, and keep track of the maximum product found.

This brute force approach works perfectly here because the constraint is small enough that checking all 2^13 - 1 = 8191 subsets is computationally feasible, and it guarantees we find the optimal answer without missing any edge cases involving negative numbers or zeros.

Learn more about Greedy, Dynamic Programming, Backtracking and Sorting patterns.

Solution Approach

The solution implements binary enumeration to explore all possible non-empty subsets of the array. Let's walk through how this works:

Binary Enumeration Setup:

• We iterate through all integers from 1 to 2^n - 1 (written as 1 << len(nums) in the code)
• Each integer i represents a unique subset where the binary representation tells us which elements to include
• Starting from 1 ensures we skip the empty subset (which would be 0 in binary)

Subset Generation Process: For each number i in our range:

1. Initialize a product variable t = 1 to accumulate the product of selected elements
2. Iterate through each element in the array using index j and value x
3. Check if the j-th bit of i is set using the operation i >> j & 1:
   • i >> j shifts the binary representation of i right by j positions
   • & 1 checks if the least significant bit is 1
   • If true, it means we include nums[j] in our current subset
4. If an element is included, multiply it into our product: t *= x

Finding the Maximum:

• We maintain a variable ans initialized to negative infinity (-inf)
• After calculating the product for each subset, we update ans with the maximum value seen so far: ans = max(ans, t)
• This ensures we track the highest product across all possible subsets

Example Walkthrough: Consider nums = [2, -3, 4]:

• When i = 1 (binary: 001), we select only nums[0] = 2, product = 2
• When i = 2 (binary: 010), we select only nums[1] = -3, product = -3
• When i = 3 (binary: 011), we select nums[0] and nums[1], product = 2 × (-3) = -6
• When i = 5 (binary: 101), we select nums[0] and nums[2], product = 2 × 4 = 8
• When i = 6 (binary: 110), we select nums[1] and nums[2], product = (-3) × 4 = -12
• When i = 7 (binary: 111), we select all elements, product = 2 × (-3) × 4 = -24

The maximum strength would be 8.

This approach guarantees we find the optimal solution by exhaustively checking all possibilities, which is feasible given the small constraint of n ≤ 13.

Example Walkthrough

Let's walk through the solution with nums = [-2, -1, 3]:

Step 1: Setup

• We have n = 3 elements, so we'll check 2^3 - 1 = 7 non-empty subsets
• We'll iterate i from 1 to 7, where each value represents a different subset

Step 2: Process each subset

When i = 1 (binary: 001):

• Check bit 0: 1 >> 0 & 1 = 1 ✓ Include nums[0] = -2
• Check bit 1: 1 >> 1 & 1 = 0 ✗ Skip nums[1]
• Check bit 2: 1 >> 2 & 1 = 0 ✗ Skip nums[2]
• Product = -2, Update ans = -2

When i = 2 (binary: 010):

• Check bit 0: 2 >> 0 & 1 = 0 ✗ Skip nums[0]
• Check bit 1: 2 >> 1 & 1 = 1 ✓ Include nums[1] = -1
• Check bit 2: 2 >> 2 & 1 = 0 ✗ Skip nums[2]
• Product = -1, Update ans = max(-2, -1) = -1

When i = 3 (binary: 011):

• Include nums[0] = -2 and nums[1] = -1
• Product = (-2) × (-1) = 2, Update ans = 2

When i = 4 (binary: 100):

• Include only nums[2] = 3
• Product = 3, Update ans = 3

When i = 5 (binary: 101):

• Include nums[0] = -2 and nums[2] = 3
• Product = (-2) × 3 = -6, Keep ans = 3

When i = 6 (binary: 110):

• Include nums[1] = -1 and nums[2] = 3
• Product = (-1) × 3 = -3, Keep ans = 3

When i = 7 (binary: 111):

• Include all elements: nums[0] = -2, nums[1] = -1, nums[2] = 3
• Product = (-2) × (-1) × 3 = 6, Update ans = 6

Step 3: Return the result The maximum strength is 6, achieved by selecting all three students.

Solution Implementation

Time and Space Complexity

The time complexity is O(2^n × n), where n is the length of the array. This is because:

• The outer loop iterates through all possible subsets of the array, from 1 to 2^n - 1 (excluding the empty subset), which gives us 2^n - 1 iterations
• For each subset represented by the bitmask i, the inner loop iterates through all n elements to check which elements belong to the current subset and multiply them together
• Therefore, the total time complexity is O(2^n × n)

The space complexity is O(1). The algorithm only uses a constant amount of extra space:

• Variable ans to store the maximum strength
• Variable t to store the product of the current subset
• Loop variables i and j
• No additional data structures that grow with input size are used

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Overflow Issues with Large Products

When multiplying many large numbers together, the product can exceed the integer limits in some programming languages. While Python handles arbitrary precision integers automatically, this could be problematic in languages like Java or C++.

Solution: In languages with fixed integer sizes, use appropriate data types (like long long in C++ or BigInteger in Java) or implement modular arithmetic if the problem requires it.

2. Inefficient Time Complexity for Larger Arrays

The binary enumeration approach has O(2^n × n) time complexity, which becomes impractical for arrays larger than ~20 elements. With n=13, we're examining 8,192 subsets, but if n=20, we'd need to check over 1 million subsets.

Solution: For larger arrays, use a greedy approach instead:

• Sort the array by absolute value in descending order
• Include all positive numbers
• Include pairs of negative numbers (starting with the largest absolute values)
• Handle edge cases with zeros and odd counts of negative numbers

def maxStrength_optimized(nums):
    if len(nums) == 1:
        return nums[0]
  
    positives = [x for x in nums if x > 0]
    negatives = [x for x in nums if x < 0]
    zeros = nums.count(0)
  
    # Sort negatives by absolute value (descending)
    negatives.sort()
  
    # If we have an even number of negatives, use all
    # If odd, exclude the one with smallest absolute value
    if len(negatives) % 2 == 1:
        negatives = negatives[:-1]
  
    # Calculate product
    if not positives and not negatives:
        return 0  # Only zeros
  
    product = 1
    for num in positives + negatives:
        product *= num
  
    return product

3. Edge Case: Single Element Array

When the array contains only one element, especially if it's negative, the code works correctly but might not be immediately obvious why.

Solution: Add explicit handling or documentation:

if len(nums) == 1:
    return nums[0]  # Only option is to select the single element

4. Memory Considerations with Subset Generation

While the current approach doesn't store all subsets in memory, developers might be tempted to generate and store all subsets first, which would use O(2^n × n) space.

Solution: Stick to the bit manipulation approach that generates subsets on-the-fly, maintaining only O(1) extra space for the product calculation.