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

Approach:



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include<bits/stdc++.h>
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
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// 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);
    }
}
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# 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
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// 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
}
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<?php
// PHP 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) 
function 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; 
  
    echo $left, " " , $right
  
// Driver code 
$row = 4; 
$col = 3; 
$x = 2; 
$y = 2; 
  
count_left_right($row, $col, $x, $y); 
  
// This code is contributed by jit_t
?>
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Output:
3 3

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