Range Sum Query 2D – Immutable LeetCode Solution

Range Sum Query 2D – Immutable 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 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)

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.

Range Sum Query 2D – Immutable Solutions

✅Time: O(mn)
✅Space: O(mn)

C++

class NumMatrix {
 public:
  NumMatrix(vector<vector<int>>& matrix) {
    if (matrix.empty())
      return;

    const int m = matrix.size();
    const int n = matrix[0].size();
    // prefix[i][j] := sum of matrix[0..i)[0..j)
    prefix.resize(m + 1, vector<int>(n + 1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        prefix[i + 1][j + 1] =
            matrix[i][j] + prefix[i][j + 1] + prefix[i + 1][j] - prefix[i][j];
  }

  int sumRegion(int row1, int col1, int row2, int col2) {
    return prefix[row2 + 1][col2 + 1] - prefix[row1][col2 + 1] -
           prefix[row2 + 1][col1] + prefix[row1][col1];
  }

 private:
  vector<vector<int>> prefix;
};

Java

 class NumMatrix {
  public NumMatrix(int[][] matrix) {
    if (matrix.length == 0)
      return;

    final int m = matrix.length;
    final int n = matrix[0].length;
    // prefix[i][j] := sum of matrix[0..i)[0..j)
    prefix = new int[m + 1][n + 1];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        prefix[i + 1][j + 1] = matrix[i][j] + prefix[i][j + 1] + prefix[i + 1][j] - prefix[i][j];
  }

  public int sumRegion(int row1, int col1, int row2, int col2) {
    return prefix[row2 + 1][col2 + 1] - prefix[row1][col2 + 1]
         - prefix[row2 + 1][col1] + prefix[row1][col1];
  }

  private int[][] prefix;
}

Python

  class NumMatrix:
  def __init__(self, matrix: List[List[int]]):
    if not matrix:
      return

    m = len(matrix)
    n = len(matrix[0])
    # prefix[i][j] := sum of matrix[0..i)[0..j)
    self.prefix = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(m):
      for j in range(n):
        self.prefix[i + 1][j + 1] = \
            matrix[i][j] + self.prefix[i][j + 1] + \
            self.prefix[i + 1][j] - self.prefix[i][j]

  def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
    return self.prefix[row2 + 1][col2 + 1] - self.prefix[row1][col2 + 1] - \
        self.prefix[row2 + 1][col1] + self.prefix[row1][col1]

Watch Tutorial

Checkout more Solutions here