Matrix Rectangular Sums - Facebook Top Interview Questions
Problem Statement :
Given a two-dimensional list of integers matrix and an integer k, return a new matrix M of the same dimensions where M[i][j] is the sum of all numbers
sum(matrix[r][c]) for all (i - k ≤ r ≤ i + k, j - k ≤ c ≤ j + k)
Constraints
n, m ≤ 250 where n and m is the number of rows and columns in matrix
Example 1
Input
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
k = 1
Output
[
[12, 21, 16],
[27, 45, 33],
[24, 39, 28]
]
Solution :
Solution in C++ :
int rectangle(vector<vector<int>>& dp, int r1, int c1, int r2, int c2) {
int res = dp[r2 + 1][c2 + 1];
res -= dp[r2 + 1][c1];
res -= dp[r1][c2 + 1];
res += dp[r1][c1];
return res;
}
vector<vector<int>> solve(vector<vector<int>>& matrix, int k) {
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> res(m, vector<int>(n));
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
dp[i + 1][j + 1] = matrix[i][j] + dp[i][j + 1] + dp[i + 1][j] - dp[i][j];
}
}
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
int r1 = max(0, r - k);
int c1 = max(0, c - k);
int r2 = min(m - 1, r + k);
int c2 = min(n - 1, c + k);
res[r][c] = rectangle(dp, r1, c1, r2, c2);
}
}
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int[][] solve(int[][] matrix, int k) {
RangeSumMatrix rsm = new RangeSumMatrix(matrix);
int[][] ret = new int[matrix.length][matrix[0].length];
for (int i = 0; i < ret.length; i++) {
for (int j = 0; j < ret[0].length; j++) {
ret[i][j] = rsm.total(i - k, j - k, i + k, j + k);
}
}
return ret;
}
}
class RangeSumMatrix {
int[][] matrix;
public RangeSumMatrix(int[][] matrix) {
this.matrix = matrix;
for (int i = 0; i < matrix.length; i++) {
for (int j = 1; j < matrix[0].length; j++) {
matrix[i][j] += matrix[i][j - 1];
}
}
}
public int total(int row0, int col0, int row1, int col1) {
row0 = Math.max(0, row0);
row1 = Math.min(matrix.length - 1, row1);
col0 = Math.max(0, col0);
col1 = Math.min(matrix[0].length - 1, col1);
int ret = 0;
for (int i = row0; i <= row1; i++) {
ret += matrix[i][col1];
if (col0 != 0) {
ret -= matrix[i][col0 - 1];
}
}
return ret;
}
}
Solution in Python :
class Solution:
def solve(self, A, k):
R, C = len(A), len(A[0])
for r in range(R):
for c in range(C):
if r:
A[r][c] += A[r - 1][c]
if c:
A[r][c] += A[r][c - 1]
if r and c:
A[r][c] -= A[r - 1][c - 1]
def get_sum(top, left, bottom, right):
top = max(top, 0)
left = max(left, 0)
bottom = min(bottom, R - 1)
right = min(right, C - 1)
return (
A[bottom][right]
- (A[top - 1][right] if top else 0)
- (A[bottom][left - 1] if left else 0)
+ (A[top - 1][left - 1] if top and left else 0)
)
return [[get_sum(r - k, c - k, r + k, c + k) for c in range(C)] for r in range(R)]
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