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Subrectangle Queries

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).

2. getValue(int row, int col)

  • Returns the current value of the coordinate (row,col) from the rectangle.

 

Example 1:

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
["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 Explanation

This problem involves designing a data structure to efficiently handle subrectangle updates and queries on a given rectangle. The solution uses a list to store the update operations. When a getValue query is made, it iterates through the update operations in reverse chronological order (from most recent to oldest). If a subrectangle update covers the queried coordinate, the updated value is returned; otherwise, the original value from the initial rectangle is used. This approach avoids the need to modify the underlying rectangle matrix directly for every update, leading to significant performance improvements for a large number of updates.

Time and Space Complexity Analysis

Time Complexity:

  • __init__ (Constructor): O(1). The constructor simply initializes the rectangle and the list to store operations. It doesn't perform any iterations proportional to the input size.

  • updateSubrectangle: O(1). The operation appends a new tuple/array to the list which takes constant time.

  • getValue: O(U), where U is the number of update operations. In the worst case, it needs to iterate through the entire list of update operations.

Space Complexity:

  • O(U), where U is the number of update operations. The space complexity is dominated by the size of the ops list, which stores the details of each update operation. The original rectangle is O(R*C), where R and C are rows and columns respectively but this is constant.

Code Explanation (Python)

The Python code efficiently implements the described approach.

class SubrectangleQueries:
    def __init__(self, rectangle: List[List[int]]):
        self.g = rectangle # Store the initial rectangle
        self.ops = []      # Initialize an empty list to store update operations
 
    def updateSubrectangle(
        self, row1: int, col1: int, row2: int, col2: int, newValue: int
    ) -> None:
        self.ops.append((row1, col1, row2, col2, newValue)) #Append update operation details
 
    def getValue(self, row: int, col: int) -> int:
        for r1, c1, r2, c2, v in self.ops[::-1]: #Iterate in reverse order of operations
            if r1 <= row <= r2 and c1 <= col <= c2: # Check if the coordinate is within the updated subrectangle
                return v #Return the updated value
        return self.g[row][col] # Otherwise return the original value

The other language implementations follow the same logic, with minor syntactic differences. The key is the use of a list to record update operations and the reverse iteration to find the most recent relevant update. This approach provides an efficient solution compared to repeatedly updating the entire rectangle.