Range Sum Query 2D – Immutable Leetcode Solution

Difficulty Level Medium
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Problem Statement

Range Sum Query 2D – Immutable Leetcode Solution – Given a 2D matrix matrix, handle multiple queries of the following type:

  • 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).

Example

Range Sum Query 2D - Immutable Leetcode Solution

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)

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

Idea

The main idea here is to store the sum for each rectangle with its top-left on (0,0) and bottom-right on (i, j) where 1<=i<=m, 1<=j<=n.

To calculate and store these sums in prefixSum matrix, we use the relation.prefixSum[i][j] = prefixSum[i-1][j] + prefixSum[i][j-1] - prefixSum[i-1][j-1] + mat[i-1][j-1];.As we are building upon the solution to the sub-problems, we are using the concept of Dynamic Programming here.

The image to imagine this relation is also attached.

Range Sum Query 2D - Immutable Leetcode Solution

Then to query each sum in O(1) time we use the relation prefixSum[r2][c2] - prefixSum[r1-1][c2] - prefixSum[r2][c1-1] + prefixSum[r1-1][c1-1];As we are building upon the solution of the sub-problems, we are using the concept of Dynamic Programming here.

The image to imagine this relation is also attached.

Range Sum Query 2D - Immutable Leetcode Solution

Code

C++ Program of  Range Sum Query 2D – Immutable

typedef vector<int> VI;
typedef vector<VI> VVI;
class NumMatrix {
private:
        int prefixSum[200+3][200+3];
public:
    NumMatrix(vector<vector<int>>& mat) {
        int n = mat.size(), m = mat[0].size();
        memset(prefixSum, 0, sizeof(prefixSum));
        
        for(int i = 1;i<=n;i++){
            for(int j = 1;j<=m;j++){
                prefixSum[i][j] = prefixSum[i-1][j] + prefixSum[i][j-1] - prefixSum[i-1][j-1] + mat[i-1][j-1];
            }
        }
    }
    
    int sumRegion(int r1, int c1, int r2, int c2) {
        r1++, c1++, c2++, r2++;
        return prefixSum[r2][c2] - prefixSum[r1-1][c2] - prefixSum[r2][c1-1] + prefixSum[r1-1][c1-1];
    }
};

Java Program of  Range Sum Query 2D – Immutable

class NumMatrix {
    long[][] prefixSum;
    
    public NumMatrix(int[][] mat) {
        int n = mat.length, m = mat[0].length;
        prefixSum = new long[n+1][m+1];
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                prefixSum[i][j] = mat[i-1][j-1] + prefixSum[i-1][j] + prefixSum[i][j-1] - prefixSum[i-1][j-1];
    }
    
    public int sumRegion(int R1, int C1, int R2, int C2) {
        R1+=1;C1+=1;R2+=1;C2+=1;
        return (int)(prefixSum[R2][C2] - prefixSum[R2][C1-1] - prefixSum[R1-1][C2] + prefixSum[R1-1][C1-1]);
    }
}

Python3 Program of  Range Sum Query 2D – Immutable

class NumMatrix:
    def __init__(self, mat: List[List[int]]):
        n, m = len(mat), len(mat[0])
        self.prefixSum = [[0] * (m+1) for _ in range(n+1)]
        for i in range(1, n+1):
            for j in range(1, m+1):
                self.prefixSum[i][j] = mat[i-1][j-1] + self.prefixSum[i-1][j] + self.prefixSum[i][j-1] - self.prefixSum[i-1][j-1]

    def sumRegion(self, R1: int, C1: int, R2: int, C2: int) -> int:
        return self.prefixSum[R2+1][C2+1] - self.prefixSum[R2+1][C1] - self.prefixSum[R1][C2+1] + self.prefixSum[R1][C1]

Complexity Analysis for Range Sum Query 2D – Immutable Leetcode Solution

Time Complexity

O(1) per query.

O(m*n) to build the prefixSum 2D matrix.

Space Complexity

O(m*n) to store the prefixSum 2D matrix.

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