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

• Last Updated : 22 Apr, 2021

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

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

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## Javascript

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Output:
`3 3`

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