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2536. Increment Submatrices by One

Difficulty Topics

Description

You are given a positive integer n, indicating that we initially have an n x n 0-indexed integer matrix mat filled with zeroes.

You are also given a 2D integer array query. For each query[i] = [row1i, col1i, row2i, col2i], you should do the following operation:

  • Add 1 to every element in the submatrix with the top left corner (row1i, col1i) and the bottom right corner (row2i, col2i). That is, add 1 to mat[x][y] for for all row1i <= x <= row2i and col1i <= y <= col2i.

Return the matrix mat after performing every query.

 

Example 1:

Input: n = 3, queries = [[1,1,2,2],[0,0,1,1]]
Output: [[1,1,0],[1,2,1],[0,1,1]]
Explanation: The diagram above shows the initial matrix, the matrix after the first query, and the matrix after the second query.
- In the first query, we add 1 to every element in the submatrix with the top left corner (1, 1) and bottom right corner (2, 2).
- In the second query, we add 1 to every element in the submatrix with the top left corner (0, 0) and bottom right corner (1, 1).

Example 2:

Input: n = 2, queries = [[0,0,1,1]]
Output: [[1,1],[1,1]]
Explanation: The diagram above shows the initial matrix and the matrix after the first query.
- In the first query we add 1 to every element in the matrix.

 

Constraints:

  • 1 <= n <= 500
  • 1 <= queries.length <= 104
  • 0 <= row1i <= row2i < n
  • 0 <= col1i <= col2i < n

Solution

increment-submatrices-by-one.py
class Solution:
    def rangeAddQueries(self, n: int, queries: List[List[int]]) -> List[List[int]]:
        A = [[0] * n for _ in range(n)]

        for r1, c1, r2, c2 in queries:
            A[r1][c1] += 1
            if c2 + 1 < n:
                A[r1][c2 + 1] -= 1
            if r2 + 1 < n:
                A[r2 + 1][c1] -= 1
            if r2 + 1 < n and c2 + 1 < n:
                A[r2 + 1][c2 + 1] += 1

        for i in range(n):
            for j in range(1, n):
                A[i][j] += A[i][j - 1]

        for i in range(1, n):
            for j in range(n):
                A[i][j] += A[i - 1][j]

        return A