Introduction

In this blog post, we will explore how to solve the LeetCode problem "1937" 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 an m x n integer matrix points (0-indexed). Starting with 0 points, you want to maximize the number of points you can get from the matrix. To gain points, you must pick one cell in each row. Picking the cell at coordinates (r, c) will add points[r][c] to your score. However, you will lose points if you pick a cell too far from the cell that you picked in the previous row. For every two adjacent rows r and r + 1 (where 0 <= r < m - 1), picking cells at coordinates (r, c1) and (r + 1, c2) will subtract abs(c1 - c2) from your score. Return the maximum number of points you can achieve. abs(x) is defined as: x for x >= 0. -x for x < 0. Example 1: Input: points = [[1,2,3],[1,5,1],[3,1,1]] Output: 9 Explanation: The blue cells denote the optimal cells to pick, which have coordinates (0, 2), (1, 1), and (2, 0). You add 3 + 5 + 3 = 11 to your score. However, you must subtract abs(2 - 1) + abs(1 - 0) = 2 from your score. Your final score is 11 - 2 = 9. Example 2: Input: points = [[1,5],[2,3],[4,2]] Output: 11 Explanation: The blue cells denote the optimal cells to pick, which have coordinates (0, 1), (1, 1), and (2, 0). You add 5 + 3 + 4 = 12 to your score. However, you must subtract abs(1 - 1) + abs(1 - 0) = 1 from your score. Your final score is 12 - 1 = 11. Constraints: m == points.length n == points[r].length 1 <= m, n <= 105 1 <= m * n <= 105 0 <= points[r][c] <= 105

Explanation

Here's a breakdown of the solution:

• Dynamic Programming: The core idea is to use dynamic programming to store the maximum points achievable at each cell.
• Optimization with Prefix/Suffix Max: Instead of recomputing the maximum score from the previous row for each cell in the current row, we use prefix and suffix maximum arrays to efficiently determine the best possible previous cell, given the penalty for distance. This avoids an O(n^2) inner loop.

• Space Optimization: The DP table is optimized to use only two rows at a time, the current row and the previous row, further reducing memory usage.

• Runtime Complexity: O(m * n)

• Storage Complexity: O(n)

Code

    def maxPoints(points):
    m = len(points)
    n = len(points[0])

    dp = [0] * n
    for j in range(n):
        dp[j] = points[0][j]

    for i in range(1, m):
        new_dp = [0] * n

        # Prefix max
        left = [0] * n
        left[0] = dp[0]
        for j in range(1, n):
            left[j] = max(left[j - 1], dp[j] + j)

        # Suffix max
        right = [0] * n
        right[n - 1] = dp[n - 1] - (n - 1)
        for j in range(n - 2, -1, -1):
            right[j] = max(right[j + 1], dp[j] - j)

        for j in range(n):
            new_dp[j] = points[i][j] + max(left[j] - j, right[j] + j)

        dp = new_dp

    return max(dp)