Dada una array de n X n . La tarea es calcular la diferencia absoluta entre las sumas de su diagonal.
Ejemplos:
Input : mat[][] = 11 2 4
4 5 6
10 8 -12
Output : 15
Sum of primary diagonal = 11 + 5 + (-12) = 4.
Sum of secondary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.
Input : mat[][] = 10 2
4 5
Output : 7
Calcular las sumas a través de las dos diagonales de una array cuadrada. A lo largo de la primera diagonal de la array, índice de fila = índice de columna, es decir, mat[i][j] se encuentra en la primera diagonal si i = j. A lo largo de la otra diagonal, índice de fila = n – 1 – índice de columna, es decir, mat[i][j] se encuentra en la segunda diagonal si i = n-1-j. Al usar dos bucles, recorremos toda la array y calculamos la suma a través de las diagonales de la array.
A continuación se muestra la implementación de este enfoque:
C++
// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#define MAX 100
using namespace std;
int difference(int arr[][MAX], int n)
{
// Initialize sums of diagonals
int d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// finding sum of primary diagonal
if (i == j)
d1 += arr[i][j];
// finding sum of secondary diagonal
if (i == n - j - 1)
d2 += arr[i][j];
}
}
// Absolute difference of the sums
// across the diagonals
return abs(d1 - d2);
}
// Driven Program
int main()
{
int n = 3;
int arr[][MAX] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
cout << difference(arr, n);
return 0;
}
Java
// JAVA Code for Find difference between sums
// of two diagonals
class GFG {
public static int difference(int arr[][], int n)
{
// Initialize sums of diagonals
int d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// finding sum of primary diagonal
if (i == j)
d1 += arr[i][j];
// finding sum of secondary diagonal
if (i == n - j - 1)
d2 += arr[i][j];
}
}
// Absolute difference of the sums
// across the diagonals
return Math.abs(d1 - d2);
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 3;
int arr[][] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
System.out.print(difference(arr, n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find the difference # between the sum of diagonal. def difference(arr, n): # Initialize sums of diagonals d1 = 0 d2 = 0 for i in range(0, n): for j in range(0, n): # finding sum of primary diagonal if (i == j): d1 += arr[i][j] # finding sum of secondary diagonal if (i == n - j - 1): d2 += arr[i][j] # Absolute difference of the sums # across the diagonals return abs(d1 - d2); # Driver Code n = 3 arr = [[11, 2, 4], [4 , 5, 6], [10, 8, -12]] print(difference(arr, n)) # This code is contributed # by ihritik
C#
// C# Code for find difference between
// sums of two diagonals
using System;
public class GFG
{
// Function to calculate difference
public static int difference(int[,] arr,
int n)
{
// Initialize sums of diagonals
int d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// finding sum of primary diagonal
if (i == j)
d1 += arr[i, j];
// finding sum of secondary diagonal
if (i == n - j - 1)
d2 += arr[i, j];
}
}
// Absolute difference of the
// sums across the diagonals
return Math.Abs(d1 - d2);
}
// Driver Code
public static void Main()
{
int n = 3;
int[,] arr ={{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}};
Console.Write(difference(arr, n));
}
}
// This code is contributed by shiv_bhakt.
PHP
<?php
// PHP program to find the difference
// between the sum of diagonal.
function difference($arr, $n)
{
// Initialize sums of diagonals
$d1 = 0; $d2 = 0;
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $n; $j++)
{
// finding sum of
// primary diagonal
if ($i == $j)
$d1 += $arr[$i][$j];
// finding sum of
// secondary diagonal
if ($i == $n - $j - 1)
$d2 += $arr[$i][$j];
}
}
// Absolute difference of the sums
// across the diagonals
return abs($d1 - $d2);
}
// Driver Code
{
$n = 3;
$arr = array(array(11, 2, 4),
array(4 , 5, 6),
array(10, 8, -12));
echo difference($arr, $n);
return 0;
}
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// Javascript Code for Find difference between sums
// of two diagonals
function difference(arr,n)
{
// Initialize sums of diagonals
let d1 = 0, d2 = 0;
for (let i = 0; i < n; i++)
{
for (let j = 0; j < n; j++)
{
// finding sum of primary diagonal
if (i == j)
d1 += arr[i][j];
// finding sum of secondary diagonal
if (i == n - j - 1)
d2 += arr[i][j];
}
}
// Absolute difference of the sums
// across the diagonals
return Math.abs(d1 - d2);
}
/* Driver program to test above function */
let n = 3;
let arr =
[
[11, 2, 4],
[4 , 5, 6],
[10, 8, -12]
];
document.write(difference(arr, n));
// This code is contributed Bobby
</script>
Producción
15
Complejidad de tiempo: O(n*n)
Espacio auxiliar: O(1) usando espacio constante para inicializar sumas diagonales
Podemos optimizar la solución anterior para que funcione en O(n) usando los patrones presentes en los índices de las celdas.
C++
// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#define MAX 100
using namespace std;
int difference(int arr[][MAX], int n)
{
// Initialize sums of diagonals
int d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
// d1 store the sum of diagonal from
// top-left to bottom-right
d1 += arr[i][i];
// d2 store the sum of diagonal from
// top-right to bottom-left
d2 += arr[i][n-i-1];
}
// Absolute difference of the sums
// across the diagonals
return abs(d1 - d2);
}
// Driven Program
int main()
{
int n = 3;
int arr[][MAX] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
cout << difference(arr, n);
return 0;
}
Java
// JAVA Code for Find difference between sums
// of two diagonals
class GFG {
public static int difference(int arr[][], int n)
{
// Initialize sums of diagonals
int d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
d1 += arr[i][i];
d2 += arr[i][n-i-1];
}
// Absolute difference of the sums
// across the diagonals
return Math.abs(d1 - d2);
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 3;
int arr[][] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
System.out.print(difference(arr, n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find the difference # between the sum of diagonal. def difference(arr, n): # Initialize sums of diagonals d1 = 0 d2 = 0 for i in range(0, n): d1 = d1 + arr[i][i] d2 = d2 + arr[i][n - i - 1] # Absolute difference of the sums # across the diagonals return abs(d1 - d2) # Driver Code n = 3 arr = [[11, 2, 4], [4 , 5, 6], [10, 8, -12]] print(difference(arr, n)) # This code is contributed # by ihritik
C#
// C# Code for find difference between
// sums of two diagonals
using System;
public class GFG
{
//Function to find difference
public static int difference(int[,] arr,
int n)
{
// Initialize sums of diagonals
int d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
d1 += arr[i, i];
d2 += arr[i, n - i - 1];
}
// Absolute difference of the sums
// across the diagonals
return Math.Abs(d1 - d2);
}
// Driver Code
public static void Main()
{
int n = 3;
int[,] arr ={{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}};
Console.Write(difference(arr, n));
}
}
// This code is contributed by shiv_bhakt.
PHP
<?php
// PHP program to find the difference
// between the sum of diagonal.
function difference($arr, $n)
{
// Initialize sums of diagonals
$d1 = 0; $d2 = 0;
for ($i = 0; $i < $n; $i++)
{
$d1 += $arr[$i][$i];
$d2 += $arr[$i][$n-$i-1];
}
// Absolute difference of the sums
// across the diagonals
return abs($d1 - $d2);
}
// Driver Code
{
$n = 3;
$arr =array(array(11, 2, 4),
array(4, 5, 6),
array(10, 8, -12));
echo difference($arr, $n);
return 0;
}
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// JAVA SCRIPT Code for Find difference between sums
// of two diagonals
function difference(arr,n)
{
// Initialize sums of diagonals
let d1 = 0, d2 = 0;
for (let i = 0; i < n; i++)
{
d1 += arr[i][i];
d2 += arr[i][n-i-1];
}
// Absolute difference of the sums
// across the diagonals
return Math.abs(d1 - d2);
}
/* Driver program to test above function */
let n = 3;
let arr = [[11, 2, 4],
[4 , 5, 6],
[10, 8, -12]];
document.write(difference(arr, n));
// This code is contributed by sravan kumar Gottumukkala
</script>
Producción
15
Complejidad de tiempo: O(n)
Espacio auxiliar: O(1 ) usando espacio constante para inicializar sumas diagonales
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA