Dados cuatro enteros fila, columna, x e y donde fila y columna son el número de filas y columnas de una array 2-D y x e y son las coordenadas de una celda en la misma array, la tarea es encontrar el número de celdas en la diagonal izquierda y derecha a la que está asociada la celda (x, y) de la array.
Ejemplos:
Entrada: fila = 4, columna = 3, x = 2, y = 2
Salida: 3 3
El número de celdas en las diagonales izquierda y derecha de (2, 2) son 3 y 3 respectivamente.
Entrada: fila = 4, columna = 5, x = 2, y = 2
Salida: 4 3
Acercarse:
- Calcule el número de celdas en la parte superior izquierda y la parte inferior derecha de la diagonal izquierda de la celda (x, y) por separado. Luego súmalos para obtener el número de celdas en la diagonal izquierda.
- Del mismo modo, calcule el número de celdas en la parte superior derecha y la parte inferior izquierda de la diagonal derecha de la celda (x, y) por separado.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
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
<?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
?>
Javascript
<script>
// Javascript 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)
{
let left = 0, right = 0;
// number of cells in the left diagonal
let left_upper_part = Math.min(x-1, y-1);
let left_lower_part = Math.min(n-x, m-y);
left = left_upper_part + left_lower_part + 1;
// number of cells in the right diagonal
let right_upper_part = Math.min(m-y, x-1);
let right_lower_part = Math.min(y-1, n-x);
right = right_upper_part + right_lower_part + 1;
document.write((left) + " " + (right));
}
// Driver code
let row = 4;
let col = 3;
let x = 2;
let y = 2;
count_left_right(row, col, x, y);
// This code is contributed by souravmahato348.
</script>
Producción:
3 3
Publicación traducida automáticamente
Artículo escrito por AmanKumarSingh y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA