Range Sum Query 2D – Immutable – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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).

You must design an algorithm where sumRegion works on O(1) time complexity.

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
• -104 <= matrix[i][j] <= 104
• 0 <= row1 <= row2 < m
• 0 <= col1 <= col2 < n
• At most 104 calls will be made to sumRegion.

C++ Range Sum Query 2D – Immutable LeetCode Solution

class NumMatrix {
private:
    int row, col;
    vector<vector<int>> sums;
public:
    NumMatrix(vector<vector<int>> &matrix) {
        row = matrix.size();
        col = row>0 ? matrix[0].size() : 0;
        sums = vector<vector<int>>(row+1, vector<int>(col+1, 0));
        for(int i=1; i<=row; i++) {
            for(int j=1; j<=col; j++) {
                sums[i][j] = matrix[i-1][j-1] + 
                             sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] ;
            }
        }
    }

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

Java Range Sum Query 2D – Immutable LeetCode Solution

private int[][] dp;

public NumMatrix(int[][] matrix) {
    if(   matrix           == null
       || matrix.length    == 0
       || matrix[0].length == 0   ){
        return;   
    }
    
    int m = matrix.length;
    int n = matrix[0].length;
    
    dp = new int[m + 1][n + 1];
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++){
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1] -dp[i - 1][j - 1] + matrix[i - 1][j - 1] ;
        }
    }
}

public int sumRegion(int row1, int col1, int row2, int col2) {
    int iMin = Math.min(row1, row2);
    int iMax = Math.max(row1, row2);
    
    int jMin = Math.min(col1, col2);
    int jMax = Math.max(col1, col2);
    
    return dp[iMax + 1][jMax + 1] - dp[iMax + 1][jMin] - dp[iMin][jMax + 1] + dp[iMin][jMin];    
}
 

Python 3 Range Sum Query 2D – Immutable LeetCode Solution

class NumMatrix(object):
      def __init__(self, matrix):
          if matrix is None or not matrix:
              return
          n, m = len(matrix), len(matrix[0])
          self.sums = [ [0 for j in xrange(m+1)] for i in xrange(n+1) ]
          for i in xrange(1, n+1):
              for j in xrange(1, m+1):
                  self.sums[i][j] = matrix[i-1][j-1] + self.sums[i][j-1] + self.sums[i-1][j] - self.sums[i-1][j-1]
    

      def sumRegion(self, row1, col1, row2, col2):
          row1, col1, row2, col2 = row1+1, col1+1, row2+1, col2+1
          return self.sums[row2][col2] - self.sums[row2][col1-1] - self.sums[row1-1][col2] + self.sums[row1-1][col1-1]

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