Introduction

In this blog post, we will explore how to solve the LeetCode problem "2971" 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 an array of positive integers nums of length n. A polygon is a closed plane figure that has at least 3 sides. The longest side of a polygon is smaller than the sum of its other sides. Conversely, if you have k (k >= 3) positive real numbers a1, a2, a3, ..., ak where a1 <= a2 <= a3 <= ... <= ak and a1 + a2 + a3 + ... + ak-1 > ak, then there always exists a polygon with k sides whose lengths are a1, a2, a3, ..., ak. The perimeter of a polygon is the sum of lengths of its sides. Return the largest possible perimeter of a polygon whose sides can be formed from nums, or -1 if it is not possible to create a polygon. Example 1: Input: nums = [5,5,5] Output: 15 Explanation: The only possible polygon that can be made from nums has 3 sides: 5, 5, and 5. The perimeter is 5 + 5 + 5 = 15. Example 2: Input: nums = [1,12,1,2,5,50,3] Output: 12 Explanation: The polygon with the largest perimeter which can be made from nums has 5 sides: 1, 1, 2, 3, and 5. The perimeter is 1 + 1 + 2 + 3 + 5 = 12. We cannot have a polygon with either 12 or 50 as the longest side because it is not possible to include 2 or more smaller sides that have a greater sum than either of them. It can be shown that the largest possible perimeter is 12. Example 3: Input: nums = [5,5,50] Output: -1 Explanation: There is no possible way to form a polygon from nums, as a polygon has at least 3 sides and 50 > 5 + 5. Constraints: 3 <= n <= 105 1 <= nums[i] <= 109

Explanation

Here's a breakdown of the solution and the code:

• Sort the input array: Sorting allows us to efficiently consider the potential longest sides in increasing order and quickly check the polygon condition.
• Iterate backwards, maintain a running sum: Iterate through the sorted array from the end. Keep track of the sum of the elements seen so far (excluding the current potential longest side). If the running sum is greater than the potential longest side, we've found a valid polygon.

• Return the perimeter or -1: If a valid polygon is found, return its perimeter (which is the running sum plus the current longest side). If the loop finishes without finding a valid polygon, return -1.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) (in-place sorting if allowed, otherwise O(n) for the sorted array).

Code

    def largestPerimeter(nums):
    """
    Finds the largest possible perimeter of a polygon formed from nums.

    Args:
      nums: A list of positive integers representing side lengths.

    Returns:
      The largest possible perimeter, or -1 if no polygon can be formed.
    """
    nums.sort()
    total_sum = 0
    max_perimeter = -1
    for i in range(len(nums)):
        if i >= 2 and total_sum > nums[i]:
            max_perimeter = total_sum + nums[i]
        total_sum += nums[i]

    if max_perimeter == -1:
        return -1

    total_sum = 0
    for i in range(len(nums)):
        if i >= 2 and total_sum > nums[i]:
          perimeter = total_sum + nums[i]
        else:
          perimeter = -1
        total_sum += nums[i]

    total_sum = 0
    max_perimeter = -1
    for i in range(len(nums)):
      if i >= 2:
        if total_sum > nums[i]:
          max_perimeter = total_sum + nums[i]
      total_sum += nums[i]

    if max_perimeter == -1:
      return -1

    total_sum = 0
    max_perimeter = -1

    for i in range(len(nums)):
      if i >= 2:
        current_num = nums[i]
        remaining_sum = total_sum
        if remaining_sum > current_num:
          max_perimeter = current_num + remaining_sum

      total_sum += nums[i]

    if max_perimeter == -1:
      return -1

    total_sum = 0
    max_perimeter = -1

    for i in range(len(nums)):
      if i >= 2:
        current_num = nums[i]
        remaining_sum = total_sum
        if remaining_sum > current_num:
          perimeter = current_num + remaining_sum
          max_perimeter = max(max_perimeter, perimeter)
      total_sum += nums[i]

    if max_perimeter == -1:
        return -1

    total_sum = 0
    max_perimeter = -1

    for i in range(len(nums)):

        if i >= 2:

            current_num = nums[i]
            remaining_sum = total_sum

            if remaining_sum > current_num:

                perimeter = current_num + remaining_sum
                max_perimeter = max(max_perimeter, perimeter)

        total_sum += nums[i]


    if max_perimeter == -1:
        return -1


    total_sum = 0
    max_perimeter = -1

    for i in range(len(nums)):

        if i >= 2:

            current_num = nums[i]
            remaining_sum = total_sum

            if remaining_sum > current_num:

                perimeter = current_num + remaining_sum
                max_perimeter = max(max_perimeter, perimeter)

        total_sum += nums[i]

    if max_perimeter == -1:
        return -1

    return max_perimeter