Problem Description

Given a matrix mat with dimensions m x n and an integer k, the task is to compute a new matrix answer of the same dimensions. For each cell answer[i][j] in the new matrix, we must calculate the sum of all elements within a square block with side length 2k + 1, centered at mat[i][j]. The boundaries of the square are constrained by the limits of the array, meaning we cannot go outside the matrix when calculating the sum. The result for answer[i][j] should only include the sum of valid elements from mat that lie within the specified range and within the matrix.

Intuition

The brute-force approach to solving this problem would involve iterating over each cell of the matrix, and for each cell, iteratively summing all the elements in the K-radius square. This would require a large number of redundant computations and would not be efficient, particularly for large matrices or values of K.

To arrive at a more efficient solution, we can use dynamic programming, specifically the concept of 2D prefix sums. The idea here is to preprocess the input matrix to generate a prefix sum matrix where each element pre[i][j] stores the sum of all elements in the rectangle defined by the upper left corner (0,0) and the lower right corner (i-1, j-1) of the original matrix. With this prefix sum matrix, we can compute the sum of any submatrix in constant time, using the inclusion-exclusion principle.

The steps to implement this are as follows:

1. Build the prefix sum matrix (pre) where the sum up to each point (i, j) is calculated by adding the current element to the sum of the elements above and to the left, subtracting the sum of the overlapping corner (already counted twice).

2. Define a helper function get(i, j) to safely retrieve the value of pre[i][j], handling boundary cases to stay within the matrix dimensions.

3. Loop through each cell (i, j) in the original matrix, and for each cell, calculate the sum of the K-radius square around it using the prefix sum matrix and the helper function. This involves adding and subtracting values from the four corners of the block surrounding (i, j), constrained by k.

The result is an efficient algorithm that computes the desired answer matrix with much fewer operations compared to the brute-force method, thus making it scalable to larger matrices and values of k. The overall time complexity is O(m * n) for the sum matrix construction plus O(m * n) for calculating the block sums, and the space complexity is O(m * n) for storing the prefix sums.

Learn more about Prefix Sum patterns.

Solution Approach

The solution to the problem uses dynamic programming with a 2D prefix sum matrix to optimize the process of computing the sums of submatrices. Let's break down the steps taken in the provided solution code:

1. Initial Prefix Sum Matrix Generation: A 2D prefix sum matrix pre is generated where pre[i][j] will contain the sum of all elements from the top-left corner (0, 0) to the position (i-1, j-1) in the original matrix mat. This matrix pre is of size (m + 1) x (n + 1) to handle the prefix sum calculation conveniently, even for the cells on the topmost row and leftmost column.

   The prefix sum is calculated as follows:

   pre[i][j] = pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1] + mat[i - 1][j - 1]

   This equation accounts for the current element and the sums above and to the left, removing the overlapped corner that was added twice.

2. Safe Prefix Sum Retrieval Function (get): The get function is defined to safely retrieve values from the pre matrix, taking care of the boundaries, and ensuring that indices outside the matrix bounds map to the nearest valid index. This function ensures that when we perform the sum calculation, we do not go outside the matrix.

   def get(i, j):
       i = max(min(m, i), 0)
       j = max(min(n, j), 0)
       return pre[i][j]

3. Computing the Block Sums: We then iterate over each cell (i, j) in the matrix mat. The sum for each answer[i][j] is computed using the get function to obtain values from the prefix sum matrix corresponding to the corners of the K-radius block:

   ans[i][j] = (
       get(i + k + 1, j + k + 1)
       - get(i + k + 1, j - k)
       - get(i - k, j + k + 1)
       + get(i - k, j - k)
   )

   This step applies the inclusion-exclusion principle to subtract the sums outside the K-radius block by considering the bottom right corner (i + k + 1, j + k + 1), the top right (i + k + 1, j - k), top left (i - k, j - k), and bottom left corners (i - k, j + k + 1).

The matrixBlockSum function ultimately returns the computed ans matrix, which represents the sum of elements within k distance from each element in the original mat.

The code structure provided makes use of a 2D prefix sum matrix and carefully handles edge cases with the get function, ensuring the solution is clean, efficient, and robust for the given problem.

Example Walkthrough

Let's use a simple 3x3 matrix as an example to illustrate how the solution approach works, with k=1:

Original matrix mat:

| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |

1. Initial Prefix Sum Matrix Generation: We first build an enlarged prefix sum matrix pre of size 4x4 to include a buffer for handling the edges. The construction would look like this after using the described formula:

   Initial pre matrix filled with zeros (for illustration):

   | 0 | 0 | 0 | 0 |
   | 0 |   |   |   |
   | 0 |   |   |   |
   | 0 |   |   |   |

   After calculating the prefix sums:

   | 0 | 0 | 0 | 0 |
   | 0 | 1 | 3 | 6 |
   | 0 | 5 |12 |21 |
   | 0 |12 |27 |45 |

   You can see that pre[3][3] has a value of 45, which is the sum of all elements of mat.

2. Safe Prefix Sum Retrieval Function (get): The get function is used to safely access elements in the pre matrix, ensuring that we do not access indices outside the bounds of the matrix.

3. Computing the Block Sums: We compute the sums for each cell in ans considering k=1. For the central element ans[1][1] corresponding to mat[1][1] which is 5, the calculation using the corners would be:

   ans[1][1] = (
       get(1 + 1 + 1, 1 + 1 + 1)
       - get(1 + 1 + 1, 1 - 1)
       - get(1 - 1, 1 + 1 + 1)
       + get(1 - 1, 1 - 1)
   )
   ans[1][1] = pre[3][3] - pre[3][1] - pre[1][3] + pre[1][1]
   ans[1][1] = 45 - 12 - 6 + 1
   ans[1][1] = 28

   The get function ensures we don't index into pre at negative indices or beyond its size by clamping the values appropriately.

   The complete ans matrix would be computed similarly, for example, the top-left corner ans[0][0] would be computed as:

   ans[0][0] = (
       get(0 + 1 + 1, 0 + 1 + 1)
       - get(0 + 1 + 1, 0 - 1)
       - get(0 - 1, 0 + 1 + 1)
       + get(0 - 1, 0 - 1)
   )
   ans[0][0] = pre[2][2] - pre[2][0] - pre[0][2] + pre[0][0]
   ans[0][0] = 12 - 0 - 0 + 0
   ans[0][0] = 12

   Which is the sum of the 2x2 block that is within k distance from the top-left element 1.

Final matrix ans after computing sums for all other cells:

| 12 | 21 | 16 |
| 27 | 45 | 33 |
| 24 | 39 | 28 |

This walk-through demonstrates the computational steps of the algorithm using the example matrix. Notice how the sums reflect the areas around each cell, constrained by k and the bounds of the matrix.

Solution Implementation

Time and Space Complexity

Time Complexity:

The time complexity of the given code can be analyzed in two major parts: the construction of the prefix sum matrix pre and the computation of the answer matrix ans.

1. Prefix Sum Matrix Computation: Constructing the prefix sum matrix requires iterating through each element of the original m x n matrix mat, and for each element, a constant amount of work is performed (adding and subtracting values). Therefore, the time complexity for this part is O(m * n).

2. Answer Matrix Computation: The computation of the answer matrix also involves iterating through all elements in the m x n matrix and performing a constant amount of work for each to calculate the sum of the block using the prefix sum matrix. Hence, the time complexity for this part is also O(m * n).

Combining both parts, the overall time complexity remains O(m * n) because both parts are sequentially executed and have the same order of complexity.

Space Complexity:

The space complexity can be analyzed as follows:

1. Prefix Sum Matrix: An additional m+1 x n+1 matrix pre is used for the prefix sums, which gives us a space complexity of O((m + 1) * (n + 1)).

2. Answer Matrix: The answer matrix ans is of the same size as the input matrix mat, i.e., m x n. Since it is used to store the final output, it might not be counted towards additional space in some analysis contexts. But for a comprehensive analysis, we consider it and, therefore, O(m * n).

When combined, the term O((m + 1) * (n + 1)) is dominant. However, since the +1 is a constant and does not change the order of growth, the overall space complexity simplifies to O(m * n).

Learn more about how to find time and space complexity quickly using problem constraints.