#1476

Medium

LeetCode Problem Workspace

Subrectangle Queries

Implement the SubrectangleQueries class to handle dynamic updates and value queries on a 2D rectangle.

Problem Statement

You are given a rows x cols rectangle represented by a 2D matrix of integers. You need to implement the SubrectangleQueries class that supports two methods:

updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) - Updates all elements within the subrectangle (row1, col1) to (row2, col2) with the given newValue.
getValue(int row, int col) - Returns the value at the specified position (row, col) in the rectangle.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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

Example 2

Input: See original problem statement.

Output: See original problem statement.

Input ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","updateSubrectangle","getValue"] [[[[1,1,1],[2,2,2],[3,3,3]]],[0,0],[0,0,2,2,100],[0,0],[2,2],[1,1,2,2,20],[2,2]] Output [null,1,null,100,100,null,20] Explanation SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,1,1],[2,2,2],[3,3,3]]); subrectangleQueries.getValue(0, 0); // return 1 subrectangleQueries.updateSubrectangle(0, 0, 2, 2, 100); subrectangleQueries.getValue(0, 0); // return 100 subrectangleQueries.getValue(2, 2); // return 100 subrectangleQueries.updateSubrectangle(1, 1, 2, 2, 20); subrectangleQueries.getValue(2, 2); // return 20

Constraints

There will be at most 500 operations considering both methods: updateSubrectangle and getValue.
1 <= rows, cols <= 100
rows == rectangle.length
cols == rectangle[i].length
0 <= row1 <= row2 < rows
0 <= col1 <= col2 < cols
1 <= newValue, rectangle[i][j] <= 10^9
0 <= row < rows
0 <= col < cols

Solution Approach

Brute Force Approach

One approach is to directly implement the updateSubrectangle method by iterating through the specified subrectangle and updating each element one by one. The getValue method can be implemented in constant time, O(1), by simply returning the value at the given coordinates.

Optimized Approach with Lazy Updates

For efficiency, we could use a lazy update technique, where updates are stored separately and applied only when necessary. This reduces the time complexity of updates while still allowing for O(1) retrieval during getValue operations.

Space Optimization

To save space, one could implement a more compact storage for updates, reducing the need to store the full matrix at all times. This approach would focus on only tracking regions that were updated and ensuring constant time for retrieval operations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The brute force approach would require O((row2 - row1 + 1) * (col2 - col1 + 1)) for each update, making it inefficient for large updates. However, the getValue operation is O(1). Optimizations can be made by reducing the number of operations through lazy updates or using efficient data structures, but they might still depend on the size of the rectangle and the updates.

What Interviewers Usually Probe

Focus on how well the candidate handles large updates in a matrix.
Look for any optimizations made to avoid redundant operations.
Test the candidate's ability to balance time and space complexity when designing the solution.

Common Pitfalls or Variants

Common pitfalls

Failing to optimize the updateSubrectangle method for larger input sizes, which may result in time inefficiencies.
Overcomplicating the solution with unnecessary space optimizations without considering the problem's constraints.
Assuming that every update is independent when they may overlap, leading to wasted operations or errors.

Follow-up variants

Allow for dynamic resizing of the rectangle after multiple updates.
Support for handling multiple different types of updates simultaneously.
Extend the functionality to support queries on entire rows or columns.

FAQ

How does the SubrectangleQueries class handle multiple updates?

The class supports multiple updates by either applying each update directly or using a lazy update approach to minimize redundant operations.

What is the time complexity of the updateSubrectangle method?

In the brute force approach, the time complexity is O((row2 - row1 + 1) * (col2 - col1 + 1)), but it can be optimized with a lazy update strategy.

Can the SubrectangleQueries class handle large matrices efficiently?

Yes, but the performance depends on the approach used. Optimizations like lazy updates can help improve efficiency when dealing with large matrices.

What is the expected space complexity of this problem?

Space complexity depends on the approach chosen, with brute force using O(m * n) space for the matrix and optimizations reducing this requirement.

How do lazy updates work in this problem?

Lazy updates work by recording changes to regions of the matrix and applying them only when necessary, reducing redundant operations and improving efficiency.

terminal

Solution

Solution 1

#### Python3

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)

Continue Practicing

#1472 design browser history #1463 cherry pickup ii #1504 count submatrices with all ones