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, 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
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