How do you solve Sum of Matrix After Queries on LeetCode?

Sum of Matrix After Queries (LeetCode #2718) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Processing queries in reverse helps identify the last effective update for each row and column.

What is the brute force solution for Sum of Matrix After Queries?

The brute force approach for Sum of Matrix After Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. We can directly apply each query to the matrix, updating the respective rows or columns. This is straightforward but inefficient for larger matrices due to the repeated updates. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Matrix After Queries?

Sum of Matrix After Queries uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Matrix After Queries?

Always clarify the problem constraints and edge cases before diving into coding. Think about the data structures that can help optimize your solution. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Sum of Matrix After Queries?

Students often forget to handle the case where a row and column are updated in the same query sequence. Not considering the last update order can lead to incorrect sums.

#2718

Sum of Matrix After Queries

Medium

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of maintaining the entire matrix, we can track the last update for each row and column. This allows us to compute the final sum efficiently without explicitly constructing the matrix.

⚙️

Algorithm

3 steps

1Step 1: Initialize two arrays to track the last value set for each row and column, and their respective types.
2Step 2: Process the queries in reverse order to determine the last effective update for each row and column.
3Step 3: Calculate the total sum by considering the last updates for each row and column, ensuring to avoid double counting.

solution.py20 lines

1def sumMatrixAfterQueries(n, queries):
2    row_values = [0] * n
3    col_values = [0] * n
4    row_last_update = [-1] * n
5    col_last_update = [-1] * n
6    for i in range(len(queries) - 1, -1, -1):
7        type_i, index_i, val_i = queries[i]
8        if type_i == 0:
9            row_values[index_i] = val_i
10            row_last_update[index_i] = i
11        else:
12            col_values[index_i] = val_i
13            col_last_update[index_i] = i
14    total_sum = 0
15    for i in range(n):
16        if row_last_update[i] > col_last_update[i]:
17            total_sum += row_values[i] * n
18        else:
19            total_sum += col_values[i] * n
20    return total_sum

ℹ

Complexity note: The time complexity is O(n) because we only traverse the queries once and then compute the sum based on the last updates, which is linear with respect to n.

1Processing queries in reverse helps identify the last effective update for each row and column.
2Using arrays to track the last values and their types avoids unnecessary matrix construction.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.