304. Range Sum Query 2D - Immutable
1. Description
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.
2. Example
Example 1:
Given matrix = [
[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]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12
3. Note
- You may assume that the matrix does not change.
- There are many calls to sumRegion function.
- You may assume that row1 ≤ row2 and col1 ≤ col2.
4. Solutions
My Accepted Solution
m = i_matrix.size(), n = i_matrix.front().size()
Time complexity : O(mn)
Space complexity : O(mn)
// we add a guard row and column so we don't need to consider the border case at 0th row or column
class NumMatrix
{
public:
NumMatrix(const vector<vector<int>> &i_matrix)
{
if(!i_matrix.empty() && !i_matrix.front().empty())
{
matrixSum = vector<vector<int>>(i_matrix.size() + 1, vector<int>(i_matrix.front().size() + 1));
for(int i = 1; i < matrixSum.size(); i++)
{
for(int j = 1; j < matrixSum.front().size(); j++)
{
matrixSum[i][j] = matrixSum[i - 1][j] + matrixSum[i][j - 1] + i_matrix[i - 1][j - 1] - matrixSum[i - 1][j - 1];
}
}
}
}
int sumRegion(int i_row1, int i_col1, int i_row2, int i_col2)
{
return matrixSum[i_row2 + 1][i_col2 + 1] - matrixSum[i_row1][i_col2 + 1] - matrixSum[i_row2 + 1][i_col1] + matrixSum[i_row1][i_col1];
}
private:
vector<vector<int>> matrixSum;
};