package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
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 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
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 * - 版权所有，转发请注明出处
 */
public class _304_RangeSumQuery2DImmutable {
    /**
     * 304. Range Sum Query 2D - Immutable
     * 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).

     Range Sum Query 2D
     The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

     Example:
     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

     time : O(m * n)
     space : O(m * n)

     */

    private int[][] sum;

    public _304_RangeSumQuery2DImmutable(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return;
        int m = matrix.length;
        int n = matrix[0].length;
        sum = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[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 sum[iMax + 1][jMax + 1] - sum[iMax + 1][jMin] - sum[iMin][jMax + 1] + sum[iMin][jMin];
    }
}