How do you solve Maximum Strength of a Group on LeetCode?

Maximum Strength of a Group (LeetCode #2708) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The product of an even number of negative numbers is positive, which can significantly increase the overall strength.

What is the time complexity of Maximum Strength of a Group?

The optimal solution for Maximum Strength of a Group runs in O(n log n) time and uses O(1) space. The time complexity is dominated by the sorting step (O(n log n)). The space complexity is constant as we only use a few variables.

What is the brute force solution for Maximum Strength of a Group?

The brute force approach for Maximum Strength of a Group is Brute Force, with O(n * 2^n) time complexity and O(1) space complexity. The brute force approach involves generating all possible non-empty subsets of the array and calculating their strengths. This method ensures we explore every combination, but it can be inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Maximum Strength of a Group?

Always clarify the problem constraints and edge cases before diving into the solution. Think about how sorting can simplify the problem, especially when dealing with products. Practice writing both brute force and optimal solutions to solidify your understanding.

What are common mistakes when solving Maximum Strength of a Group?

Not considering the case of odd negative numbers and how to handle them. Forgetting to check if there are any positive numbers before returning the maximum negative.

#2708

Maximum Strength of a Group

Medium

ArrayDynamic ProgrammingBacktrackingGreedyBit ManipulationSortingEnumerationSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * 2^n)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution leverages sorting and the properties of multiplication. By sorting the array, we can efficiently determine the best combination of positive and negative numbers to maximize the product.

⚙️

Algorithm

5 steps

1Step 1: Sort the array in non-decreasing order.
2Step 2: Initialize a product variable to 1 and a count of negative numbers to 0.
3Step 3: Iterate through the sorted array, multiplying positive numbers and counting negative numbers.
4Step 4: If there are an odd number of negative numbers, exclude the least negative number from the product.
5Step 5: If no positive numbers exist, return the maximum single negative number.

solution.py15 lines

1def maxStrength(nums):
2    nums.sort()
3    product = 1
4    negative_count = 0
5    max_negative = float('-inf')
6    for num in nums:
7        if num > 0:
8            product *= num
9        elif num < 0:
10            product *= num
11            negative_count += 1
12            max_negative = max(max_negative, num)
13    if negative_count % 2 == 1:
14        product //= max_negative
15    return product if product != 1 else max(max_negative, 0)

ℹ

Complexity note: The time complexity is dominated by the sorting step (O(n log n)). The space complexity is constant as we only use a few variables.

1The product of an even number of negative numbers is positive, which can significantly increase the overall strength.
2Including all positive numbers is always beneficial as they increase the product.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.