Leetcode #1314: Matrix Block Sum
In this guide, we solve Leetcode #1314 Matrix Block Sum in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a m x n matrix mat and an integer k, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for: i - k <= r <= i + k, j - k <= c <= j + k, and (r, c) is a valid position in the matrix. Example 1: Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 1 Output: [[12,21,16],[27,45,33],[24,39,28]] Example 2: Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 2 Output: [[45,45,45],[45,45,45],[45,45,45]] Constraints: m == mat.length n == mat[i].length 1 <= m, n, k <= 100 1 <= mat[i][j] <= 100
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Matrix, Prefix Sum
Intuition
Range queries become simple once we precompute cumulative sums.
We can transform subarray conditions into prefix comparisons.
Approach
Compute prefix sums and use a map to find matching prefixes.
This avoids nested loops while keeping the logic clear.
Steps:
- Compute prefix sums.
- Use a map to find valid ranges.
- Update the answer.
Example
Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]
Python Solution
class Solution:
def matrixBlockSum(self, mat: List[List[int]], k: int) -> List[List[int]]:
m, n = len(mat), len(mat[0])
s = [[0] * (n + 1) for _ in range(m + 1)]
for i, row in enumerate(mat, 1):
for j, x in enumerate(row, 1):
s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x
ans = [[0] * n for _ in range(m)]
for i in range(m):
for j in range(n):
x1, y1 = max(i - k, 0), max(j - k, 0)
x2, y2 = min(m - 1, i + k), min(n - 1, j + k)
ans[i][j] = (
s[x2 + 1][y2 + 1] - s[x1][y2 + 1] - s[x2 + 1][y1] + s[x1][y1]
)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.