How do you solve Matrix Block Sum on LeetCode?

Matrix Block Sum (LeetCode #1314) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Using prefix sums allows for efficient submatrix sum calculations.

What is the brute force solution for Matrix Block Sum?

The brute force approach for Matrix Block Sum is Brute Force, with O(m * n * (2k + 1)²) time complexity and O(1) space complexity. For each cell in the matrix, we will calculate the sum of all elements within the k-distance. This is straightforward but can be slow, especially for larger matrices. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Matrix Block Sum?

Matrix Block Sum uses the following concepts: Array, Matrix, Prefix Sum. The recommended patterns to study are: Prefix Sum, Matrix Manipulation.

What are the interview tips for Matrix Block Sum?

Explain your thought process clearly. Discuss the trade-offs between brute force and optimal solutions. Practice writing clean and efficient code.

What are common mistakes when solving Matrix Block Sum?

Not handling matrix boundaries correctly. Forgetting to initialize the prefix sum matrix.

#1314

Matrix Block Sum

Medium

ArrayMatrixPrefix SumPrefix SumMatrix Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * (2k + 1)²)	O(m * n)
Space	O(1)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

Using a prefix sum array allows us to calculate the sum of any submatrix efficiently, reducing the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum matrix where each cell (i, j) contains the sum of all elements from (0, 0) to (i, j).
2Step 2: For each cell (i, j) in the result matrix, calculate the sum of the submatrix defined by (i-k, j-k) to (i+k, j+k) using the prefix sum matrix.
3Step 3: Handle boundaries to ensure we do not go out of bounds.

solution.py15 lines

1# Full working Python code
2
3def matrixBlockSum(mat, k):
4    m, n = len(mat), len(mat[0])
5    prefix = [[0] * (n + 1) for _ in range(m + 1)]
6    for i in range(m):
7        for j in range(n):
8            prefix[i + 1][j + 1] = mat[i][j] + prefix[i][j + 1] + prefix[i + 1][j] - prefix[i][j]
9    result = [[0] * n for _ in range(m)]
10    for i in range(m):
11        for j in range(n):
12            r1, c1 = max(0, i - k), max(0, j - k)
13            r2, c2 = min(m - 1, i + k), min(n - 1, j + k)
14            result[i][j] = prefix[r2 + 1][c2 + 1] - prefix[r1][c2 + 1] - prefix[r2 + 1][c1] + prefix[r1][c1]
15    return result

ℹ

Complexity note: The time complexity is reduced to O(m * n) because we compute the prefix sums in a single pass, and then each cell's sum can be computed in constant time. The space complexity is O(m * n) due to the prefix sum matrix.

1Using prefix sums allows for efficient submatrix sum calculations.
2Understanding boundaries is crucial when working with matrices.

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