# Solving Leetcode Interviews in Seconds with AI: Subrectangle Queries


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1476" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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).  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  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Data Storage:** Store the original rectangle (matrix) directly.
*   **Update Operation:** Iterate through the specified subrectangle and update each cell with the new value.
*   **Get Value Operation:** Directly access the matrix at the specified row and column to return the value.

*   **Runtime Complexity:** `updateSubrectangle` is O(m*n) where m and n are the dimensions of the subrectangle and `getValue` is O(1).
    **Storage Complexity:** O(rows * cols)

	
	# Code
	```python
	class SubrectangleQueries:

    def __init__(self, rectangle: list[list[int]]):
        self.rectangle = rectangle

    def updateSubrectangle(self, row1: int, col1: int, row2: int, col2: int, newValue: int) -> None:
        for i in range(row1, row2 + 1):
            for j in range(col1, col2 + 1):
                self.rectangle[i][j] = newValue

    def getValue(self, row: int, col: int) -> int:
        return self.rectangle[row][col]
	```