Leetcode #1476: Subrectangle Queries
In this guide, we solve Leetcode #1476 Subrectangle Queries 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
Implement the class SubrectangleQueries which receives a rows x cols rectangle as a matrix of integers in the constructor and supports two methods: 1. updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) Updates all values with newValue in the subrectangle whose upper left coordinate is (row1,col1) and bottom right coordinate is (row2,col2).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Array, Matrix
Intuition
Grid problems are easiest when you define clear row/column boundaries.
A consistent traversal order prevents off-by-one errors.
Approach
Iterate by rows, columns, or layers depending on the requirement.
Keep bounds updated as the traversal progresses.
Steps:
- Define bounds or directions.
- Visit cells in order.
- Update result and move bounds.
Example
Input
["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","updateSubrectangle","getValue","getValue"]
[[[[1,2,1],[4,3,4],[3,2,1],[1,1,1]]],[0,2],[0,0,3,2,5],[0,2],[3,1],[3,0,3,2,10],[3,1],[0,2]]
Output
[null,1,null,5,5,null,10,5]
Explanation
SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,2,1],[4,3,4],[3,2,1],[1,1,1]]);
// The initial rectangle (4x3) looks like:
// 1 2 1
// 4 3 4
// 3 2 1
// 1 1 1
subrectangleQueries.getValue(0, 2); // return 1
subrectangleQueries.updateSubrectangle(0, 0, 3, 2, 5);
// After this update the rectangle looks like:
// 5 5 5
// 5 5 5
// 5 5 5
// 5 5 5
subrectangleQueries.getValue(0, 2); // return 5
subrectangleQueries.getValue(3, 1); // return 5
subrectangleQueries.updateSubrectangle(3, 0, 3, 2, 10);
// After this update the rectangle looks like:
// 5 5 5
// 5 5 5
// 5 5 5
// 10 10 10
subrectangleQueries.getValue(3, 1); // return 10
subrectangleQueries.getValue(0, 2); // return 5
Python Solution
class SubrectangleQueries:
def __init__(self, rectangle: List[List[int]]):
self.g = rectangle
self.ops = []
def updateSubrectangle(
self, row1: int, col1: int, row2: int, col2: int, newValue: int
) -> None:
self.ops.append((row1, col1, row2, col2, newValue))
def getValue(self, row: int, col: int) -> int:
for r1, c1, r2, c2, v in self.ops[::-1]:
if r1 <= row <= r2 and c1 <= col <= c2:
return v
return self.g[row][col]
# Your SubrectangleQueries object will be instantiated and called as such:
# obj = SubrectangleQueries(rectangle)
# obj.updateSubrectangle(row1,col1,row2,col2,newValue)
# param_2 = obj.getValue(row,col)
Complexity
The time complexity is O(m·n). The space complexity is O(1) to O(m·n).
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.