Dada una array cuadrada 2D, encuentre la suma de los elementos en las diagonales Principal y Secundaria. Por ejemplo, considere la siguiente array de entrada de 4 X 4.
A00 A01 A02 A03 A10 A11 A12 A13 A20 A21 A22 A23 A30 A31 A32 A33
La diagonal primaria está formada por los elementos A00, A11, A22, A33.
- Condición para la Diagonal Principal: La condición fila-columna es fila = columna.
La diagonal secundaria está formada por los elementos A03, A12, A21, A30. - Condición para la diagonal secundaria: la condición fila-columna es fila = número de filas – columna -1.
Ejemplos:
Input : 4 1 2 3 4 4 3 2 1 7 8 9 6 6 5 4 3 Output : Principal Diagonal: 16 Secondary Diagonal: 20 Input : 3 1 1 1 1 1 1 1 1 1 Output : Principal Diagonal: 3 Secondary Diagonal: 3
Método 1: en este método, usamos dos bucles, es decir, un bucle para columnas y un bucle para filas y en el bucle interno verificamos la condición establecida anteriormente:
Implementación:
C++
// A simple C++ program to find sum of diagonals
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
void printDiagonalSums(int mat[][MAX], int n)
{
int principal = 0, secondary = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for principal diagonal
if (i == j)
principal += mat[i][j];
// Condition for secondary diagonal
if ((i + j) == (n - 1))
secondary += mat[i][j];
}
}
cout << "Principal Diagonal:" << principal << endl;
cout << "Secondary Diagonal:" << secondary << endl;
}
// Driver code
int main()
{
int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
{ 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
return 0;
}
Java
// A simple java program to find
// sum of diagonals
import java.io.*;
public class GFG {
static void printDiagonalSums(int [][]mat,
int n)
{
int principal = 0, secondary = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for principal
// diagonal
if (i == j)
principal += mat[i][j];
// Condition for secondary
// diagonal
if ((i + j) == (n - 1))
secondary += mat[i][j];
}
}
System.out.println("Principal Diagonal:"
+ principal);
System.out.println("Secondary Diagonal:"
+ secondary);
}
// Driver code
static public void main (String[] args)
{
int [][]a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
}
}
// This code is contributed by vt_m.
Python3
# A simple Python program to
# find sum of diagonals
MAX = 100
def printDiagonalSums(mat, n):
principal = 0
secondary = 0;
for i in range(0, n):
for j in range(0, n):
# Condition for principal diagonal
if (i == j):
principal += mat[i][j]
# Condition for secondary diagonal
if ((i + j) == (n - 1)):
secondary += mat[i][j]
print("Principal Diagonal:", principal)
print("Secondary Diagonal:", secondary)
# Driver code
a = [[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ]]
printDiagonalSums(a, 4)
# This code is contributed
# by ihritik
C#
// A simple C# program to find sum
// of diagonals
using System;
public class GFG {
static void printDiagonalSums(int [,]mat,
int n)
{
int principal = 0, secondary = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for principal
// diagonal
if (i == j)
principal += mat[i,j];
// Condition for secondary
// diagonal
if ((i + j) == (n - 1))
secondary += mat[i,j];
}
}
Console.WriteLine("Principal Diagonal:"
+ principal);
Console.WriteLine("Secondary Diagonal:"
+ secondary);
}
// Driver code
static public void Main ()
{
int [,]a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
}
}
// This code is contributed by vt_m.
PHP
<?php
// A simple PHP program to
// find sum of diagonals
$MAX = 100;
function printDiagonalSums($mat, $n)
{
global $MAX;
$principal = 0;
$secondary = 0;
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $n; $j++)
{
// Condition for
// principal diagonal
if ($i == $j)
$principal += $mat[$i][$j];
// Condition for
// secondary diagonal
if (($i + $j) == ($n - 1))
$secondary += $mat[$i][$j];
}
}
echo "Principal Diagonal:" ,
$principal ,"\n";
echo "Secondary Diagonal:",
$secondary ,"\n";
}
// Driver code
$a = array (array ( 1, 2, 3, 4 ),
array ( 5, 6, 7, 8 ),
array ( 1, 2, 3, 4 ),
array ( 5, 6, 7, 8 ));
printDiagonalSums($a, 4);
// This code is contributed by ajit
?>
Javascript
<script>
// A simple Javascript program to find sum of diagonals
const MAX = 100;
void printDiagonalSums(mat, n)
{
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
// Condition for principal diagonal
if (i == j)
principal += mat[i][j];
// Condition for secondary diagonal
if ((i + j) == (n - 1))
secondary += mat[i][j];
}
}
document.write("Principal Diagonal:" + principal + "<br>");
document.write("Secondary Diagonal:" + secondary + "<br>");
}
// Driver code
let a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ];
printDiagonalSums(a, 4);
// This code is contributed by subhammahato348.
</script>
Producción
Principal Diagonal:18 Secondary Diagonal:18
Complejidad de tiempo : O(N*N) , ya que estamos usando bucles anidados para atravesar N*N veces.
Espacio auxiliar : O(1), ya que no estamos utilizando ningún espacio adicional.
Método 2 (enfoque eficiente): en este método usamos un bucle, es decir, un bucle para calcular la suma de las diagonales principal y secundaria:
Implementación:
C++
// An efficient C++ program to find sum of diagonals
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
void printDiagonalSums(int mat[][MAX], int n)
{
int principal = 0, secondary = 0;
for (int i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
cout << "Principal Diagonal:" << principal << endl;
cout << "Secondary Diagonal:" << secondary << endl;
}
// Driver code
int main()
{
int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
{ 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
return 0;
}
Java
// An efficient java program to find
// sum of diagonals
import java.io.*;
public class GFG {
static void printDiagonalSums(int [][]mat,
int n)
{
int principal = 0, secondary = 0;
for (int i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
System.out.println("Principal Diagonal:"
+ principal);
System.out.println("Secondary Diagonal:"
+ secondary);
}
// Driver code
static public void main (String[] args)
{
int [][]a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
}
}
// This code is contributed by vt_m.
Python3
# A simple Python3 program to find
# sum of diagonals
MAX = 100
def printDiagonalSums(mat, n):
principal = 0
secondary = 0
for i in range(0, n):
principal += mat[i][i]
secondary += mat[i][n - i - 1]
print("Principal Diagonal:", principal)
print("Secondary Diagonal:", secondary)
# Driver code
a = [[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ]]
printDiagonalSums(a, 4)
# This code is contributed
# by ihritik
C#
// An efficient C#program to find
// sum of diagonals
using System;
public class GFG {
static void printDiagonalSums(int [,]mat,
int n)
{
int principal = 0, secondary = 0;
for (int i = 0; i < n; i++) {
principal += mat[i,i];
secondary += mat[i,n - i - 1];
}
Console.WriteLine("Principal Diagonal:"
+ principal);
Console.WriteLine("Secondary Diagonal:"
+ secondary);
}
// Driver code
static public void Main ()
{
int [,]a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
}
}
// This code is contributed by vt_m.
PHP
<?php
// An efficient PHP program
// to find sum of diagonals
$MAX = 100;
function printDiagonalSums($mat, $n)
{
global $MAX;
$principal = 0; $secondary = 0;
for ($i = 0; $i < $n; $i++)
{
$principal += $mat[$i][$i];
$secondary += $mat[$i][$n - $i - 1];
}
echo "Principal Diagonal:" ,
$principal ,"\n";
echo "Secondary Diagonal:" ,
$secondary ,"\n";
}
// Driver Code
$a = array(array(1, 2, 3, 4),
array(5, 6, 7, 8),
array(1, 2, 3, 4),
array(5, 6, 7, 8));
printDiagonalSums($a, 4);
// This code is contributed by aj_36
?>
Javascript
<script>
// An efficient Javascript program to find
// sum of diagonals
function printDiagonalSums(mat,n)
{
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
document.write("Principal Diagonal:"
+ principal+"<br>");
document.write("Secondary Diagonal:"
+ secondary);
}
// Driver code
let a = [[ 1, 2, 3, 4 ],
[5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ]];
printDiagonalSums(a, 4);
// This code is contributed Bobby
</script>
Producción
Principal Diagonal:18 Secondary Diagonal:18
Complejidad de tiempo: O (N), ya que estamos usando un bucle para atravesar N veces.
Espacio auxiliar: O(1) , ya que no estamos utilizando ningún espacio adicional.
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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