Range Sum Query 2D - Mutable

Medium

Asked by:

11 views

Topics:

ArraysDynamic Programming

Given a 2D matrix matrix, handle multiple queries of the following type:

Update the value of a cell in matrix.
Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Implement the NumMatrix class:

NumMatrix(int[][] matrix) Initializes the object with the 2D integer matrix matrix.
void update(int row, int col, int val) Updates the value of matrix[row][col] to be val.
int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

You must design an algorithm where update and sumRegion are both O(log(n)) time complexity.

Example 1:

Input
["NumMatrix", "sumRegion", "update", "sumRegion"]
[[[[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]], [2, 1, 4, 3], [3, 2, 2], [2, 1, 4, 3]]
Output
[null, 8, null, 10]

Explanation
NumMatrix numMatrix = new NumMatrix([[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8
numMatrix.update(3, 2, 2);
numMatrix.sumRegion(2, 1, 4, 3); // return 10

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-10⁵ <= matrix[i][j] <= 10⁵
0 <= row < m
0 <= col < n
-10⁵ <= val <= 10⁵
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
At most 500 calls will be made to update and sumRegion.

Try coding it in LeetCode

Range Sum Query 2D - Mutable

Medium

Asked by:

11 views

Topics:

ArraysDynamic Programming

Given a 2D matrix matrix, handle multiple queries of the following type:

Update the value of a cell in matrix.
Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Implement the NumMatrix class:

NumMatrix(int[][] matrix) Initializes the object with the 2D integer matrix matrix.
void update(int row, int col, int val) Updates the value of matrix[row][col] to be val.
int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

You must design an algorithm where update and sumRegion are both O(log(n)) time complexity.

Example 1:

Input
["NumMatrix", "sumRegion", "update", "sumRegion"]
[[[[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]], [2, 1, 4, 3], [3, 2, 2], [2, 1, 4, 3]]
Output
[null, 8, null, 10]

Explanation
NumMatrix numMatrix = new NumMatrix([[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8
numMatrix.update(3, 2, 2);
numMatrix.sumRegion(2, 1, 4, 3); // return 10

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-10⁵ <= matrix[i][j] <= 10⁵
0 <= row < m
0 <= col < n
-10⁵ <= val <= 10⁵
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
At most 500 calls will be made to update and sumRegion.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the matrix (number of rows and columns), and what is the maximum size these dimensions can be?
Can the values within the matrix be negative, zero, or floating-point numbers, or are they strictly positive integers?
How frequently will the `update` and `sumRegion` methods be called relative to the size of the matrix? Is there a particular ratio or pattern I should optimize for?
If `row1 > row2` or `col1 > col2` in the `sumRegion` method, what should be the expected behavior/return value?
Is the input matrix guaranteed to be rectangular, or can it be ragged/irregular?

Brute Force Solution

Approach

The brute force method calculates range sums by directly recomputing the sum within the specified region every time it's needed. When the matrix changes, we simply update the matrix with the new value and move on. Basically, it's doing the most obvious and straightforward calculation each time.

Here's how the algorithm would work step-by-step:

For calculating the sum of a region, go through each row that is part of the desired region.
Within each row, go through each number that is part of the desired region.
Add up all those numbers.
The final sum is the answer.
For updating a value in the matrix, simply find the number you need to change and replace it with the new value. Nothing else needs to change until a sum is requested again.

Code Implementation

class NumMatrix:
    def __init__(self, matrix):
        self.matrix = matrix

    def update(self, row, col, value):
        # Updates the value in the matrix at the specified position

        self.matrix[row][col] = value

    def sumRegion(self, row1, col1, row2, col2):
        total_sum = 0

        # Iterate through the rows within the specified range
        for row_index in range(row1, row2 + 1):

            # Iterate through the columns within the specified range
            for col_index in range(col1, col2 + 1):
                total_sum += self.matrix[row_index][col_index]

        return total_sum

Big(O) Analysis

Time Complexity

O(m*n) – The sumRegion operation iterates through the rows and columns within the specified rectangular region of the matrix. In the worst case, the region can cover the entire matrix, where 'm' represents the number of rows and 'n' represents the number of columns. Therefore, the operation requires iterating through all m rows and within each row, iterating through all n columns, leading to m*n operations. The update operation simply updates a single element which takes O(1) time but the sum operation remains O(m*n) in the worst case. Thus the dominant cost of querying is O(m*n).

Space Complexity

O(1) – The brute force method, as described, performs calculations directly on the input matrix and updates it in place. No auxiliary data structures like temporary arrays, hash maps, or recursion stacks are used. Therefore, the algorithm uses a constant amount of extra space regardless of the size of the input matrix, which we can denote as N. The auxiliary space is independent of N, so it is O(1).

Optimal Solution

Approach

The key to efficiently solving this problem lies in pre-calculating and storing sums of rectangular regions of the matrix. This allows us to quickly retrieve range sums and update values without recalculating the whole matrix repeatedly. We use a structure that is similar to the original matrix, but helps with these two operations.

Here's how the algorithm would work step-by-step:

Create a new structure that has the same dimensions as the original matrix.
The value at each position in this new structure will represent the sum of all values in the original matrix from the top left corner of the original matrix to the current position.
When asked to calculate the sum of a rectangular region, use the precalculated sums in this new structure to determine the sum using some addition and subtraction of the precalculated values.
When asked to update a value in the original matrix, update the corresponding value in the original matrix first.
After updating the original matrix, also update all the corresponding values in the structure. Only the sums related to the changed value need to be updated.
By precalculating and storing these sums, we avoid recomputing the sums every time we need them.

Code Implementation

class NumMatrix:

    def __init__(self, matrix):
        self.matrix = matrix
        if not matrix or not matrix[0]:
            self.sum_matrix = []
            return

        rows = len(matrix)
        cols = len(matrix[0])
        self.sum_matrix = [[0] * (cols + 1) for _ in range(rows + 1)]

        # Precalculate the sum of all values from the top left.
        for row_index in range(1, rows + 1):
            for col_index in range(1, cols + 1):
                self.sum_matrix[row_index][col_index] = self.sum_matrix[row_index - 1][col_index] + self.sum_matrix[row_index][col_index - 1] - self.sum_matrix[row_index - 1][col_index - 1] + matrix[row_index - 1][col_index - 1]

    def update(self, row, col, value):
        # Updates the original matrix with the new value.
        self.matrix[row][col] = value
        rows = len(self.matrix)
        cols = len(self.matrix[0])
        self.sum_matrix = [[0] * (cols + 1) for _ in range(rows + 1)]

        # Recalculate the sum matrix after the update.
        for row_index in range(1, rows + 1):
            for col_index in range(1, cols + 1):
                self.sum_matrix[row_index][col_index] = self.sum_matrix[row_index - 1][col_index] + self.sum_matrix[row_index][col_index - 1] - self.sum_matrix[row_index - 1][col_index - 1] + self.matrix[row_index - 1][col_index - 1]

    def sumRegion(self, row1, col1, row2, col2):
        # Use precalculated sums to determine the region sum.
        return self.sum_matrix[row2 + 1][col2 + 1] - self.sum_matrix[row1][col2 + 1] - self.sum_matrix[row2 + 1][col1] + self.sum_matrix[row1][col1]

Big(O) Analysis

Time Complexity

O(m*n) – Constructing the auxiliary matrix, where each element stores the sum of the rectangle from the top-left corner, requires iterating through each cell of the original m x n matrix. Thus the initialization takes O(m*n) time. Updating a single element in the original matrix would require updating the auxiliary matrix. The update of the auxiliary matrix requires at most updating all the sums related to the changed value which can potentially be all the entries after the updated element. The sumRange query uses constant time because it involves looking up precalculated values and doing some arithmetic operations.

Space Complexity

O(N*M) – The solution creates a new data structure (prefix sum matrix) with the same dimensions as the original matrix, where N is the number of rows and M is the number of columns in the input matrix. This new structure stores precalculated sums, requiring N*M space. The additional space used is directly proportional to the size of the original matrix. Therefore, the auxiliary space complexity is O(N*M).

Edge Cases

Case	How to Handle
Null or empty matrix	Return immediately from constructor or other methods if matrix is null or has zero rows or columns, to prevent NullPointerExceptions.
Update indices out of matrix bounds	Throw an IllegalArgumentException or ignore the update if row or col index is out of bounds.
Sum range indices out of matrix bounds	Throw an IllegalArgumentException or adjust indices to the valid range if row1, col1, row2, or col2 is out of bounds.
row1 > row2 or col1 > col2 in sumRegion	Swap row1 and row2 or col1 and col2 to ensure that row1 <= row2 and col1 <= col2 before computing the sum.
Integer overflow in sumRegion calculation	Use long to store the intermediate sum during the sumRegion calculation to avoid integer overflow.
Very large matrix dimensions (close to integer limits)	Ensure the data structure and algorithm used scales efficiently to handle matrices with dimensions close to integer limits, potentially impacting memory usage or runtime complexity.
Matrix with very large or very small numbers (close to integer limits)	Consider using long or double if the matrix elements can be very large or very small, to avoid overflow or precision issues during updates or sum calculations.
Numerous update operations followed by a single sumRegion operation	Consider optimizing the update operation to be efficient as excessive updates with infrequent queries could degrade overall performance depending on the selected algorithm