Introduction

In this blog post, we will explore how to solve the LeetCode problem "2280" 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 integer array stockPrices where stockPrices[i] = [dayi, pricei] indicates the price of the stock on day dayi is pricei. A line chart is created from the array by plotting the points on an XY plane with the X-axis representing the day and the Y-axis representing the price and connecting adjacent points. One such example is shown below: Return the minimum number of lines needed to represent the line chart. Example 1: Input: stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]] Output: 3 Explanation: The diagram above represents the input, with the X-axis representing the day and Y-axis representing the price. The following 3 lines can be drawn to represent the line chart: - Line 1 (in red) from (1,7) to (4,4) passing through (1,7), (2,6), (3,5), and (4,4). - Line 2 (in blue) from (4,4) to (5,4). - Line 3 (in green) from (5,4) to (8,1) passing through (5,4), (6,3), (7,2), and (8,1). It can be shown that it is not possible to represent the line chart using less than 3 lines. Example 2: Input: stockPrices = [[3,4],[1,2],[7,8],[2,3]] Output: 1 Explanation: As shown in the diagram above, the line chart can be represented with a single line. Constraints: 1 <= stockPrices.length <= 105 stockPrices[i].length == 2 1 <= dayi, pricei <= 109 All dayi are distinct.

Explanation

Here's a breakdown of the solution:

• Sort by Day: Sort the stockPrices array by the day (the first element of each inner array) to ensure the points are in chronological order.
• Calculate Slope: Iterate through the sorted points, calculating the slope between consecutive points.

• Count Line Changes: Increment the line count whenever the slope changes.

• Time Complexity: O(n log n) due to sorting.

• Space Complexity: O(1) excluding the space used for sorting (which might be O(log n) or O(n) depending on the sorting algorithm).

Code

    def minimumLines(stockPrices):
    stockPrices.sort()
    n = len(stockPrices)
    if n <= 1:
        return 0

    lines = 1
    def slope(p1, p2):
        return (p2[1] - p1[1]) / (p2[0] - p1[0])

    prev_slope = slope(stockPrices[1], stockPrices[0])

    for i in range(2, n):
        curr_slope = slope(stockPrices[i], stockPrices[i-1])
        if curr_slope != prev_slope:
            lines += 1
            prev_slope = curr_slope

    return lines