Introduction

In this blog post, we will explore how to solve the LeetCode problem "2500" 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 matrix grid consisting of positive integers. Perform the following operation until grid becomes empty: Delete the element with the greatest value from each row. If multiple such elements exist, delete any of them. Add the maximum of deleted elements to the answer. Note that the number of columns decreases by one after each operation. Return the answer after performing the operations described above. Example 1: Input: grid = [[1,2,4],[3,3,1]] Output: 8 Explanation: The diagram above shows the removed values in each step. - In the first operation, we remove 4 from the first row and 3 from the second row (notice that, there are two cells with value 3 and we can remove any of them). We add 4 to the answer. - In the second operation, we remove 2 from the first row and 3 from the second row. We add 3 to the answer. - In the third operation, we remove 1 from the first row and 1 from the second row. We add 1 to the answer. The final answer = 4 + 3 + 1 = 8. Example 2: Input: grid = [[10]] Output: 10 Explanation: The diagram above shows the removed values in each step. - In the first operation, we remove 10 from the first row. We add 10 to the answer. The final answer = 10. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 50 1 <= grid[i][j] <= 100

Explanation

Here's the breakdown of the solution:

• Sort Rows: Sort each row of the matrix in descending order. This ensures that the largest element in each row is always at the front.
• Iterate and Find Maximum: Iterate through the columns (from left to right after sorting). In each column, find the maximum element among all rows.

• Accumulate Result: Add the maximum element found in each column to the running answer.

• Runtime Complexity: O(m n log(n)) due to sorting each row.

• Storage Complexity: O(1) (in-place sorting). If in-place sorting is not allowed then O(n) because a copy of each row needs to be created.

Code

    def deleteGreatestValue(grid):
    """
    Calculates the sum of the maximum deleted values from each column of the grid.

    Args:
        grid: A list of lists of integers representing the grid.

    Returns:
        The sum of the maximum deleted values.
    """
    m = len(grid)
    n = len(grid[0])
    ans = 0

    for i in range(m):
        grid[i].sort(reverse=True)

    for j in range(n):
        max_val = 0
        for i in range(m):
            max_val = max(max_val, grid[i][j])
        ans += max_val

    return ans