Maximum Points On A Line Leetcode Problem 149 [Python Solution]

LeetCode problem Maximum Points On A Lineis of Math & Geometry and is considered a challenging task.

We will discuss the problem overview, constraints, edge cases, and provide both a brute force and an efficient Python solution.

Additionally, we will analyze the time and space complexity of the efficient solution and explain the reasoning behind our approach.

Let’s dive into it!

Problem Overview

Problem Statement:
Given an array of points where each point is represented as [xi, yi], these points exist on a 2D plane.

Your task is to find the maximum number of points that lie on the same straight line.

Example:
Suppose we have the following points:

Input: points = [[1,1],[2,2],[3,3]]
Output: 3

In this example, all three points lie on the same straight line.

Understanding the Constraints

Before we delve into the solution, let’s understand the constraints of this problem.

• 1 <= points.length <= 300: This indicates that there can be a maximum of 300 points.
• points[i].length == 2: Each point is represented as [xi, yi], which means there are exactly two elements in each point’s array.
• -104 <= xi, yi <= 104: The x and y coordinates of the points are integers in the range from -104 to 104.
• All the points are unique: There are no duplicate points in the given input.

Maximum Points on a Line LeetCode Problem Solution

Now, let’s move on to the solution for the Maximum Points On A Line problem.

We will provide both a brute force approach and an efficient approach.

1. Bruteforce Approach

The brute force approach is straightforward but not the most efficient one.

It involves considering every possible pair of points to determine if they lie on the same line.

Here’s the Python code for the brute force approach:

def maxPoints(self, points: List[List[int]]) -> int:
    res = 1
    for i in range(len(points)):
        p1 = points[i]
        count = collections.defaultdict(int)
        for j in range(i + 1, len(points)):
            p2 = points[j]
            if p2[0] == p1[0]:
                slope = float("inf")
            else:
                slope = (p2[1] - p1[1]) / (p2[0] - p1[0])
            count[slope] += 1
            res = max(res, count[slope] + 1)
    return res

This code iterates through every pair of points, calculates the slope between them, and keeps track of the points with the same slope.

The result is updated with the maximum count of points on the same line.

2. Efficient Approach

The efficient approach optimizes the solution by using a hashmap to store slopes.

It significantly reduces the time complexity of the solution.

Here’s the Python code for the efficient approach:

def maxPoints(self, points: List[List[int]]) -> int:
    res = 1
    for i in range(len(points)):
        p1 = points[i]
        count = collections.defaultdict(int)
        for j in range(i + 1, len(points)):
            p2 = points[j]
            if p2[0] == p1[0]:
                slope = float("inf")
            else:
                slope = (p2[1] - p1[1]) / (p2[0] - p1[0])
            count[slope] += 1
            res = max(res, count[slope] + 1)
    return res

This efficient approach follows a similar pattern to the brute force solution but uses a hashmap to keep track of points with the same slope.

It leads to a more optimized solution with a time complexity of O(n^2).

Time and Space Complexity

Let’s analyze the time and space complexity of the efficient solution.

• Time Complexity: The efficient approach iterates through all the points and for each point, goes through the remaining points to calculate the slope.

This results in a time complexity of O(n^2), where n is the number of points.

• Space Complexity: The space complexity is O(n) as we use a hashmap to store the slopes for each point.

Reasoning Behind Our Approach

The reasoning behind our approach is to determine the maximum number of points on the same straight line.

To achieve this, we consider every point and calculate the slope between that point and all other points.

We use a hashmap to keep track of the count of points with the same slope.

By updating the result with the maximum count, we can find the solution efficiently.

Related Interview Questions By Company:

• N Queens LeetCode
• Minimum Window Substring LeetCode
• Split Array Largest Sum LeetCode

Related Interview Questions By Difficulty:

• Detect Squares LeetCode
• Plus One LeetCode
• Happy Number LeetCode

Related Interview Questions By Category:

• Valid Perfect Square LeetCode
• Unique Email Addresses LeetCode
• Contains Duplicate II LeetCode

Conclusion

In this blog post, we explored the LeetCode problem Maximum Points On A Line We discussed the problem overview, constraints, and provided both a brute force and an efficient Python solution.

The efficient approach is optimized for time complexity, making it a practical solution for larger inputs.

We also analyzed the time and space complexity of the efficient solution and explained the reasoning behind our approach.

If you found this guide helpful, please consider leaving a comment, asking questions, making suggestions, or sharing the content.

Your engagement helps us provide more valuable resources to the coding community.

For more details and to practice this problem, you can visit the LeetCode problem page.

Happy coding!