304. Range Sum Query 2D - Immutable

Description

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

1. 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 integer matrix matrix.

• 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).

Constraints

• m == matrix.length

• n == matrix[i].length

• 1 <= m, n <= 200

• -105 <= matrix[i][j] <= 105

• 0 <= row1 <= row2 < m

• 0 <= col1 <= col2 < n

• At most 104 calls will be made to sumRegion.

Approach

Algorithm

We used a cumulative sum array in the 1D version. We notice that the cumulative sum is computed with respect to the origin at index 0. Extending this analogy to the 2D case, we could pre-compute a cumulative region sum with respect to the origin at (0, 0).

Sum(OD) is the cumulative region sum with respect to the origin at (0, 0).

How do we derive Sum(ABCD)Sum(ABCD) using the pre-computed cumulative region sum?

Sum(OB) is the cumulative region sum on top of the rectangle.

Sum(OC) is the cumulative region sum to the left of the rectangle.

Sum(OA) is the cumulative region sum to the top left corner of the rectangle.

Note that the region Sum(OA) is covered twice by both Sum(OB) and Sum(OC). We could use the principle of inclusion-exclusion to calculate Sum(ABCD) as following:

Sum(ABCD) = Sum(OD) − Sum(OB) − Sum(OC) + Sum(OA)

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input:

["NumMatrix", "sumRegion", "sumRegion", "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], [1, 1, 2, 2], [1, 2, 2, 4]]

Output:

[null, 8, 11, 12]

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 (i.e sum of the red rectangle)

numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)

numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

Solutions

Solution 1

/**
 * Time complexity : O(1) time per query, O(mn) time pre-computation. 
 *    The pre-computation in the constructor takes O(mn) time. 
 *    Each sumRegion query takes O(1) time.
 * Space complexity : O(mn). The algorithm uses O(mn) space to store 
 *    the cumulative region sum.
 */

class NumMatrix {
    private int[][] dp;

    public NumMatrix(int[][] matrix) {
        int rows = matrix.length, cols = matrix[0].length;
        dp = new int[rows+1][cols+1];
        for(int i = 1; i < rows+1; i++) {
            for(int j = 1; j < cols+1; j++) {
                dp[i][j] = matrix[i-1][j-1] + dp[i][j-1] + dp[i-1][j] - dp[i-1][j-1];
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        return dp[row2+1][col2+1] - dp[row2+1][col1] - dp[row1][col2+1] + dp[row1][col1];
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

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