Introduction

In this blog post, we will explore how to solve the LeetCode problem "2708" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• Handle Zeros: Remove all zeros from the input array. If the array becomes empty after removing zeros, and the original array had zeros, return 0. Otherwise, proceed with the remaining elements.
  • Maximize Positive Product: If there are positive numbers, multiply all of them. If there are an even number of negative numbers, multiply them as well. If there are an odd number of negative numbers, exclude the smallest (least negative) negative number to maximize the product.
  • Handle All Negatives: If there are no positive numbers, and only negative numbers remain, return the product of the two largest (least negative) numbers if there are at least two numbers. If only one negative number remains, return that number.

• Runtime Complexity: O(n log n) due to sorting (if sorting is needed), otherwise O(n). Storage Complexity: O(1) - constant extra space.

Code

    def maxStrength(nums):
    """
    Calculates the maximum strength of a group of students.

    Args:
        nums: A list of integers representing student scores.

    Returns:
        The maximum strength of a group.
    """
    positives = []
    negatives = []
    zeros = 0

    for num in nums:
        if num > 0:
            positives.append(num)
        elif num < 0:
            negatives.append(num)
        else:
            zeros += 1

    if not positives and not negatives and zeros > 0:
        return 0

    if not positives and not negatives:
        return 0  # Handle case where nums is all zeros (already handled above). This line is redundant, but makes the intent clear.

    product = 1
    for num in positives:
        product *= num

    if len(negatives) % 2 == 0:
        for num in negatives:
            product *= num
    elif len(negatives) > 0:
        negatives.sort()
        for i in range(len(negatives) - 1):
            product *= negatives[i]

        if not positives and len(negatives) == 1 and zeros == 0:
            return negatives[-1]
        elif not positives and len(negatives) == 1 and zeros > 0:
             return 0 # the logic handled it correctly, but it won't reach here


    if product == 1 and not positives and len(negatives) == 1:  # Only one negative number.
        return negatives[0]

    if product == 1 and not positives and len(negatives) == 0 and zeros > 0 :
        return 0

    if product == 1 and not positives and len(negatives) == 0 and zeros == 0:
        return 1 #This case won't happen

    if product == 1 and len(negatives) == 0 and len(positives) == 0 and zeros > 0 :
        return 0

    if product == 1 and not positives and len(negatives) ==1 and zeros == 0:
      return negatives[0]

    return product