You are given a 2D boolean matrix grid.
Return an integer that is the number of right triangles that can be made with the 3 elements of grid such that all of them have a value of 1.
Note:
\n\ngrid is a right triangle if one of its elements is in the same row with another element and in the same column with the third element. The 3 elements do not have to be next to each other.\n
Example 1:
\n\n| 0 | \n\t\t\t1 | \n\t\t\t0 | \n\t\t
| 0 | \n\t\t\t1 | \n\t\t\t1 | \n\t\t
| 0 | \n\t\t\t1 | \n\t\t\t0 | \n\t\t
| 0 | \n\t\t\t1 | \n\t\t\t0 | \n\t\t
| 0 | \n\t\t\t1 | \n\t\t\t1 | \n\t\t
| 0 | \n\t\t\t1 | \n\t\t\t0 | \n\t\t
Input: grid = [[0,1,0],[0,1,1],[0,1,0]]
\n\nOutput: 2
\n\nExplanation:
\n\nThere are two right triangles.
\nExample 2:
\n\n| 1 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t
| 0 | \n\t\t\t1 | \n\t\t\t0 | \n\t\t\t1 | \n\t\t
| 1 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t
Input: grid = [[1,0,0,0],[0,1,0,1],[1,0,0,0]]
\n\nOutput: 0
\n\nExplanation:
\n\nThere are no right triangles.
\nExample 3:
\n\n| 1 | \n\t\t\t0 | \n\t\t\t1 | \n\t\t
| 1 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t
| 1 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t
| 1 | \n\t\t\t0 | \n\t\t\t1 | \n\t\t
| 1 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t
| 1 | \n\t\t\t0 | \n\t\t\t0 | \n\t\t
Input: grid = [[1,0,1],[1,0,0],[1,0,0]]
\n\nOutput: 2
\n\nExplanation:
\n\nThere are two right triangles.
\n\n
Constraints:
\n\n1 <= grid.length <= 10001 <= grid[i].length <= 10000 <= grid[i][j] <= 1grid[x][y] is 1, it can form a right triangle with an element of grid with value 1 in the same row and an element of grid with value 1 in the same column.",
"So we just need to count the number of 1s in each row and column.",
"For each x, y with grid[x][y] = 1 if there are row[x] 1s in the row x and col[y] 1s in column y, the answer should be added by (row[x] - 1) * (col[y] - 1)."
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