Introduction

In this blog post, we will explore how to solve the LeetCode problem "2718" 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 integer n and a 0-indexed 2D array queries where queries[i] = [typei, indexi, vali]. Initially, there is a 0-indexed n x n matrix filled with 0's. For each query, you must apply one of the following changes: if typei == 0, set the values in the row with indexi to vali, overwriting any previous values. if typei == 1, set the values in the column with indexi to vali, overwriting any previous values. Return the sum of integers in the matrix after all queries are applied. Example 1: Input: n = 3, queries = [[0,0,1],[1,2,2],[0,2,3],[1,0,4]] Output: 23 Explanation: The image above describes the matrix after each query. The sum of the matrix after all queries are applied is 23. Example 2: Input: n = 3, queries = [[0,0,4],[0,1,2],[1,0,1],[0,2,3],[1,2,1]] Output: 17 Explanation: The image above describes the matrix after each query. The sum of the matrix after all queries are applied is 17. Constraints: 1 <= n <= 104 1 <= queries.length <= 5 * 104 queries[i].length == 3 0 <= typei <= 1 0 <= indexi < n 0 <= vali <= 105

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Instead of maintaining a full matrix, we'll use two arrays (rows and cols) to track the last value assigned to each row and column, along with the query index at which it was assigned.
  • Iterating through the queries in reverse order allows us to prioritize later queries (overwriting earlier ones).
  • Calculate the sum by iterating through the conceptual matrix, using the row and column arrays to determine the value of each cell.

• Complexity:

  • Runtime: O(n + q), where n is the dimension of the matrix and q is the number of queries.
  • Storage: O(n) due to the rows and cols arrays.

Code

    def matrix_sum_after_queries(n: int, queries: list[list[int]]) -> int:
    """
    Calculates the sum of integers in the matrix after applying all queries.

    Args:
        n: The dimension of the matrix (n x n).
        queries: A list of queries, where each query is a list [type, index, val].

    Returns:
        The sum of integers in the matrix after all queries are applied.
    """

    rows = {}  # row_index: (value, query_index)
    cols = {}  # col_index: (value, query_index)

    matrix_sum = 0

    for query_index in range(len(queries) - 1, -1, -1):
        type_i, index_i, val_i = queries[query_index]

        if type_i == 0:  # Row update
            if index_i not in rows:
                rows[index_i] = (val_i, query_index)
        else:  # Column update
            if index_i not in cols:
                cols[index_i] = (val_i, query_index)

    for r in range(n):
        for c in range(n):
            row_val, row_query_index = rows.get(r, (0, -1))
            col_val, col_query_index = cols.get(c, (0, -1))

            if row_query_index > col_query_index:
                matrix_sum += row_val
            else:
                matrix_sum += col_val

    return matrix_sum