Matrix Block Sum

Given a m * n matrix mat and an integer K, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for i - K <= r <= i + K, j - K <= c <= j + K, and (r, c) is a valid position in the matrix.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]

Example 2:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]

Constraints:

• m == mat.length

• n == mat[i].length

• 1 <= m, n, K <= 100

• 1 <= mat[i][j] <= 100

class Solution {
    public int[][] matrixBlockSum(int[][] mat, int K) {
        int m = mat.length, n = mat[0].length;
        int prefixSum[][] = new int[m][n];
        for (int i = 0; i < m; i++) {
            int sum = 0;
            for (int j = 0; j < n; j++) {
                sum += mat[i][j];
                if (i == 0)
                    prefixSum[i][j] = sum;
                else
                    prefixSum[i][j] = sum + prefixSum[i - 1][j];
            }
        }
        int[][] ans = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int rUP = i + K, cUP = j + K;
                if (rUP >= m)
                    rUP = m - 1;
                if (cUP >= n)
                    cUP = n - 1;
                ans[i][j] = prefixSum[rUP][cUP];
                int r = i - K - 1;
                int c = j - K - 1;
                if (r < 0 && c < 0)
                    continue;
                else if (r < 0)
                    ans[i][j] = ans[i][j] - prefixSum[rUP][c];

                else if (c < 0)
                    ans[i][j] = ans[i][j] - prefixSum[r][cUP];
                else
                    ans[i][j] = ans[i][j] - prefixSum[rUP][c] - prefixSum[r][cUP] + prefixSum[r][c];
            }
        }
        return ans;
    }
}