# Solving Leetcode Interviews in Seconds with AI: Find the Number of Ways to Place People I


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3025" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 2D array points of size n x 2 representing integer coordinates of some points on a 2D plane, where points[i] = [xi, yi]. Count the number of pairs of points (A, B), where  A is on the upper left side of B, and there are no other points in the rectangle (or line) they make (including the border).  Return the count.   Example 1:  Input: points = [[1,1],[2,2],[3,3]] Output: 0 Explanation:  There is no way to choose A and B so A is on the upper left side of B.  Example 2:  Input: points = [[6,2],[4,4],[2,6]] Output: 2 Explanation:   The left one is the pair (points[1], points[0]), where points[1] is on the upper left side of points[0] and the rectangle is empty. The middle one is the pair (points[2], points[1]), same as the left one it is a valid pair. The right one is the pair (points[2], points[0]), where points[2] is on the upper left side of points[0], but points[1] is inside the rectangle so it's not a valid pair.   Example 3:  Input: points = [[3,1],[1,3],[1,1]] Output: 2 Explanation:   The left one is the pair (points[2], points[0]), where points[2] is on the upper left side of points[0] and there are no other points on the line they form. Note that it is a valid state when the two points form a line. The middle one is the pair (points[1], points[2]), it is a valid pair same as the left one. The right one is the pair (points[1], points[0]), it is not a valid pair as points[2] is on the border of the rectangle.     Constraints:  2 <= n <= 50 points[i].length == 2 0 <= points[i][0], points[i][1] <= 50 All points[i] are distinct.  

	# Explanation
	Here's a breakdown of the solution:

*   **Iterate and Check:** The core idea is to iterate through all possible pairs of points (A, B) and check if A is in the upper-left region of B.
*   **Rectangle Emptiness:** For each valid pair (A, B), we iterate through all other points to ensure that no other point lies within or on the boundary of the rectangle formed by A and B.
*   **Counting Valid Pairs:** If a pair satisfies both conditions (A is upper-left of B, and the rectangle is empty), increment the count.

*   **Runtime Complexity:** O(n^3), where n is the number of points.  **Storage Complexity:** O(1)

	
	# Code
	```python
	def count_valid_pairs(points):
    """
    Counts the number of pairs of points (A, B) where A is on the upper left side of B,
    and there are no other points in the rectangle they make (including the border).

    Args:
        points: A list of lists of integers, where points[i] = [xi, yi].

    Returns:
        The count of valid pairs.
    """

    n = len(points)
    count = 0

    for i in range(n):
        for j in range(n):
            if i == j:
                continue

            a_x, a_y = points[i]
            b_x, b_y = points[j]

            if a_x < b_x and a_y < b_y:
                is_valid = True
                for k in range(n):
                    if k == i or k == j:
                        continue

                    c_x, c_y = points[k]

                    if a_x <= c_x <= b_x and a_y <= c_y <= b_y:
                        is_valid = False
                        break

                if is_valid:
                    count += 1

    return count
	```