Introduction

In this blog post, we will explore how to solve the LeetCode problem "3462" 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 matrix grid of size n x m, an integer array limits of length n, and an integer k. The task is to find the maximum sum of at most k elements from the matrix grid such that: The number of elements taken from the ith row of grid does not exceed limits[i]. Return the maximum sum. Example 1: Input: grid = [[1,2],[3,4]], limits = [1,2], k = 2 Output: 7 Explanation: From the second row, we can take at most 2 elements. The elements taken are 4 and 3. The maximum possible sum of at most 2 selected elements is 4 + 3 = 7. Example 2: Input: grid = [[5,3,7],[8,2,6]], limits = [2,2], k = 3 Output: 21 Explanation: From the first row, we can take at most 2 elements. The element taken is 7. From the second row, we can take at most 2 elements. The elements taken are 8 and 6. The maximum possible sum of at most 3 selected elements is 7 + 8 + 6 = 21. Constraints: n == grid.length == limits.length m == grid[i].length 1 <= n, m <= 500 0 <= grid[i][j] <= 105 0 <= limits[i] <= m 0 <= k <= min(n * m, sum(limits))

Explanation

Here's the breakdown of the approach and the Python code:

• High-Level Approach:

  • For each row, sort the elements in descending order. This allows us to greedily pick the largest elements up to the limit for that row.
  • Use dynamic programming to determine the maximum sum achievable by considering the first i rows and selecting a total of j elements.
  • The DP table dp[i][j] stores the maximum sum achievable considering the first i rows and taking exactly j elements in total.

• Complexity:

  • Runtime Complexity: O(n * m * k), where n is the number of rows, m is the number of columns, and k is the maximum number of elements allowed.
  • Storage Complexity: O(n * k) due to the DP table.

Code

    def max_sum(grid, limits, k):
    """
    Calculates the maximum sum of at most k elements from the grid,
    respecting the limits for each row.

    Args:
        grid: A 2D integer matrix.
        limits: An integer array of limits for each row.
        k: The maximum number of elements to select.

    Returns:
        The maximum sum achievable.
    """

    n = len(grid)
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        row = sorted(grid[i - 1], reverse=True)
        limit = limits[i - 1]
        for j in range(k + 1):
            dp[i][j] = dp[i - 1][j]  # Don't take anything from current row
            current_sum = 0
            for x in range(min(j, limit)):  # Take 1 to limit elements from the row
                current_sum += row[x]
                dp[i][j] = max(dp[i][j], dp[i - 1][j - (x + 1)] + current_sum)

    return dp[n][k]