# Number of cells in the right and left diagonals passing through (x, y) in a matrix

Given four integers row, col, x and y where row and col are the number of rows and columns of a 2-D Matrix and x and y are the coordinates of a cell in the same matrix, the task is to find number of cells in the left and the right diagonal which the cell (x, y) of the matrix is associated with.

Examples:

Input: row = 4, col = 3, x = 2, y = 2
Output: 3 3

The number of cells in the left and the right diagonals of (2, 2) are 3 and 3 respectively.

Input: row = 4, col = 5, x = 2, y = 2
Output: 4 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Calculate the number of cells in the upper left part and lower right part of the left diagonal of the cell (x, y) separately. Then sum them up to get the number of cells in the left diagonal.
• Similarly, calculate the number of cells in the upper right part and lower left part of the right diagonal of the cell (x, y) separately.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `    ``// Function to return the number of cells ` `    ``// in the left and the right diagonal of ` `    ``// the matrix for a cell (x, y) ` `    ``void` `count_left_right(``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `    ``{ ` `        ``int` `left = 0, right = 0; ` ` `  `        ``// number of cells in the left diagonal ` `        ``int` `left_upper_part = min(x-1, y-1); ` `        ``int` `left_lower_part = min(n-x, m-y); ` `        ``left = left_upper_part + left_lower_part + 1; ` ` `  `        ``// number of cells in the right diagonal ` `        ``int` `right_upper_part = min(m-y, x-1); ` `        ``int` `right_lower_part = min(y-1, n-x); ` `        ``right = right_upper_part + right_lower_part + 1; ` ` `  `        ``cout<<(left)<<``" "``<<(right); ` `    ``} ` ` `  `    ``// Driver code ` `    ``int` `main() ` `    ``{ ` `        ``int` `row = 4; ` `        ``int` `col = 3; ` `        ``int` `x = 2; ` `        ``int` `y = 2; ` ` `  `        ``count_left_right(row, col, x, y); ` `    ``} ` `// This code is contributed by  ` `// Sanjit_Prasad `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG { ` ` `  `    ``// Function to return the number of cells ` `    ``// in the left and the right diagonal of ` `    ``// the matrix for a cell (x, y) ` `    ``static` `void` `count_left_right(``int` `n, ``int` `m ` `                                    ``, ``int` `x, ``int` `y) ` `    ``{ ` `        ``int` `left = ``0``, right = ``0``; ` ` `  `        ``// number of cells in the left diagonal ` `        ``int` `left_upper_part = Math.min(x - ``1``, y - ``1``); ` `        ``int` `left_lower_part = Math.min(n - x, m - y); ` `        ``left = left_upper_part + left_lower_part + ``1``; ` ` `  `        ``// number of cells in the right diagonal ` `        ``int` `right_upper_part = Math.min(m - y, x - ``1``); ` `        ``int` `right_lower_part = Math.min(y - ``1``, n - x); ` `        ``right = right_upper_part + right_lower_part + ``1``; ` ` `  `        ``System.out.println(left + ``" "` `+ right); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `row = ``4``; ` `        ``int` `col = ``3``; ` `        ``int` `x = ``2``; ` `        ``int` `y = ``2``; ` ` `  `        ``count_left_right(row, col, x, y); ` `    ``} ` `} `

## Python 3

 `# Python 3 implementation of the approach ` ` `  `# Function to return the number of cells ` `# in the left and the right diagonal of ` `# the matrix for a cell (x, y) ` `def` `count_left_right(n, m, x, y): ` `     `  `    ``left ``=` `0` `    ``right ``=` `0` ` `  `    ``# number of cells in the left diagonal ` `    ``left_upper_part ``=` `min``(x ``-` `1``, y ``-` `1``) ` `    ``left_lower_part ``=` `min``(n ``-` `x, m ``-` `y) ` `    ``left ``=` `left_upper_part ``+` `left_lower_part ``+` `1` ` `  `    ``# number of cells in the right diagonal ` `    ``right_upper_part ``=` `min``(m ``-` `y, x ``-` `1``) ` `    ``right_lower_part ``=` `min``(y ``-` `1``, n ``-` `x) ` `    ``right ``=` `right_upper_part ``+` `right_lower_part ``+` `1` ` `  `    ``print``(left, right) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``row ``=` `4` `    ``col ``=` `3` `    ``x ``=` `2` `    ``y ``=` `2` ` `  `    ``count_left_right(row, col, x, y) ` ` `  `# This code is contributed by ChitraNayal `

## C#

 `// C# implementation of the above approach ` ` `  `using` `System; ` ` `  `class` `Program ` `{ ` `    ``// Function to return the number of cells ` `    ``// in the left and the right diagonal of ` `    ``// the matrix for a cell (x, y) ` `    ``static` `void` `count_left_right(``int` `n, ``int` `m ` `                            ``, ``int` `x, ``int` `y) ` `    ``{ ` `        ``int` `left = 0, right = 0; ` `         `  `        ``// number of cells in the left diagonal ` `        ``int` `left_upper_part = Math.Min(x - 1, y - 1); ` `        ``int` `left_lower_part = Math.Min(n - x, m - y); ` `        ``left = left_upper_part + left_lower_part + 1; ` `         `  `        ``// number of cells in the right diagonal ` `        ``int` `right_upper_part = Math.Min(m - y, x - 1); ` `        ``int` `right_lower_part = Math.Min(y - 1, n - x); ` `        ``right = right_upper_part + right_lower_part + 1; ` `        ``Console.WriteLine(left + ``" "` `+ right); ` `    ``} ` `     `  `    ``//Driver code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``int` `row = 4; ` `        ``int` `col = 3; ` `        ``int` `x = 2; ` `        ``int` `y = 2; ` `        ``count_left_right(row, col, x, y); ` `         `  `    ``} ` `// This code is contributed by ANKITRAI1 ` `} `

## PHP

 ` `

Output:

```3 3
```

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.