Introduction

In this blog post, we will explore how to solve the LeetCode problem "611" 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

Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle. Example 1: Input: nums = [2,2,3,4] Output: 3 Explanation: Valid combinations are: 2,3,4 (using the first 2) 2,3,4 (using the second 2) 2,2,3 Example 2: Input: nums = [4,2,3,4] Output: 4 Constraints: 1 <= nums.length <= 1000 0 <= nums[i] <= 1000

Explanation

Here's an efficient solution to count the number of possible triangles from an array of side lengths:

• Sort the array: Sorting allows us to efficiently check the triangle inequality. For a triangle to be valid with sides a, b, and c, the sum of any two sides must be greater than the third side (a + b > c, a + c > b, b + c > a). If we sort, and assume a <= b <= c, we only need to check a + b > c.
• Two-pointer approach: After sorting, iterate through the array from the end (largest side 'c'). Use two pointers, 'left' and 'right', starting from the beginning and the element before 'c' respectively. Move the pointers based on whether nums[left] + nums[right] > nums[c].

• Count valid triplets: If nums[left] + nums[right] > nums[c], it means all elements between 'left' and 'right' will also satisfy the triangle inequality with nums[right] and nums[c]. So, we increment the count by (right - left) and decrement 'right'. Otherwise, we increment 'left'.

• Runtime Complexity: O(n^2), where n is the length of the input array. Sorting takes O(n log n), but the nested loop dominates.

• Storage Complexity: O(1). The algorithm operates in place (after sorting) and uses a constant amount of extra space.

Code

    def triangleNumber(nums):
    """
    Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
    """
    nums.sort()
    count = 0
    n = len(nums)

    for i in range(n - 1, 0, -1):
        left = 0
        right = i - 1

        while (left < right):
            if (nums[left] + nums[right] > nums[i]):
                count += (right - left)
                right -= 1
            else:
                left += 1

    return count