Dada una array cuadrada 2D, imprima las diagonales Principal y Secundaria.
Ejemplos:
Input: 4 1 2 3 4 4 3 2 1 7 8 9 6 6 5 4 3 Output: Principal Diagonal: 1, 3, 9, 3 Secondary Diagonal: 4, 2, 8, 6 Input: 3 1 1 1 1 1 1 1 1 1 Output: Principal Diagonal: 1, 1, 1 Secondary Diagonal: 1, 1, 1
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 Diagonal Principal:
The row-column condition is row = column.
- La diagonal secundaria está formada por los elementos A03, A12, A21, A30.
Condición para la diagonal secundaria:
The row-column condition is row = numberOfRows - column -1.
Método 1:
En este método, usamos dos bucles, es decir, un bucle para columnas y un bucle para filas y en el bucle interior verificamos la condición indicada anteriormente.
C++
// C++ Program to print the Diagonals of a Matrix
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
// Function to print the Principal Diagonal
void printPrincipalDiagonal(int mat[][MAX], int n)
{
cout << "Principal Diagonal: ";
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for principal diagonal
if (i == j)
cout << mat[i][j] << ", ";
}
}
cout << endl;
}
// Function to print the Secondary Diagonal
void printSecondaryDiagonal(int mat[][MAX], int n)
{
cout << "Secondary Diagonal: ";
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for secondary diagonal
if ((i + j) == (n - 1))
cout << mat[i][j] << ", ";
}
}
cout << endl;
}
// Driver code
int main()
{
int n = 4;
int a[][MAX] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
return 0;
}
Java
// Java Program to print the Diagonals of a Matrix
class GFG {
static int MAX = 100;
// Function to print the Principal Diagonal
static void printPrincipalDiagonal(int mat[][], int n)
{
System.out.print("Principal Diagonal: ");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for principal diagonal
if (i == j) {
System.out.print(mat[i][j] + ", ");
}
}
}
System.out.println("");
}
// Function to print the Secondary Diagonal
static void printSecondaryDiagonal(int mat[][], int n)
{
System.out.print("Secondary Diagonal: ");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for secondary diagonal
if ((i + j) == (n - 1)) {
System.out.print(mat[i][j] + ", ");
}
}
}
System.out.println("");
}
// Driver code
public static void main(String args[])
{
int n = 4;
int a[][] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 Program to print the Diagonals of a Matrix
MAX = 100
# Function to print the Principal Diagonal
def printPrincipalDiagonal(mat, n):
print("Principal Diagonal: ", end = "")
for i in range(n):
for j in range(n):
# Condition for principal diagonal
if (i == j):
print(mat[i][j], end = ", ")
print()
# Function to print the Secondary Diagonal
def printSecondaryDiagonal(mat, n):
print("Secondary Diagonal: ", end = "")
for i in range(n):
for j in range(n):
# Condition for secondary diagonal
if ((i + j) == (n - 1)):
print(mat[i][j], end = ", ")
print()
# Driver code
n = 4
a = [[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ]]
printPrincipalDiagonal(a, n)
printSecondaryDiagonal(a, n)
# This code is contributed by Mohit Kumar
C#
// C# Program to print the Diagonals of a Matrix
using System;
class GFG {
static int MAX = 100;
// Function to print the Principal Diagonal
static void printPrincipalDiagonal(int[, ] mat, int n)
{
Console.Write("Principal Diagonal: ");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for principal diagonal
if (i == j) {
Console.Write(mat[i, j] + ", ");
}
}
}
Console.WriteLine("");
}
// Function to print the Secondary Diagonal
static void printSecondaryDiagonal(int[, ] mat, int n)
{
Console.Write("Secondary Diagonal: ");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Condition for secondary diagonal
if ((i + j) == (n - 1)) {
Console.Write(mat[i, j] + ", ");
}
}
}
Console.WriteLine("");
}
// Driver code
public static void Main(String[] args)
{
int n = 4;
int[, ] a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript Program to print the Diagonals of a Matrix
let MAX = 100;
// Function to print the Principal Diagonal
function printPrincipalDiagonal(mat, n)
{
document.write("Principal Diagonal: ");
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
// Condition for principal diagonal
if (i == j) {
document.write(mat[i][j] + ", ");
}
}
}
document.write("</br>");
}
// Function to print the Secondary Diagonal
function printSecondaryDiagonal(mat, n)
{
document.write("Secondary Diagonal: ");
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
// Condition for secondary diagonal
if ((i + j) == (n - 1)) {
document.write(mat[i][j] + ", ");
}
}
}
document.write("</br>");
}
let n = 4;
let a = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ] ];
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
</script>
Producción:
Principal Diagonal: 1, 6, 3, 8, Secondary Diagonal: 4, 7, 2, 5,
Análisis de Complejidad:
- Complejidad Temporal: O(n 2 ).
Como hay un bucle anidado involucrado, la complejidad del tiempo se eleva al cuadrado. - Espacio Auxiliar: O(1).
Como no se ocupa espacio extra.
Método 2:
en este método, se puede lograr la misma condición para imprimir los elementos diagonales usando un solo bucle for.
Acercarse:
- Para los elementos de la Diagonal Principal: Ejecute un bucle for a hasta n , donde n es el número de columnas , e imprima array[i][i] donde i es la variable de índice.
- Para los elementos de la diagonal secundaria: Ejecute un ciclo hasta n , donde n es el número de columnas e imprima array[i][k] donde i es la variable de índice y k = array_length – 1 . Disminuir k hasta i < n .
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ Program to print the Diagonals of a Matrix
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
// Function to print the Principal Diagonal
void printPrincipalDiagonal(int mat[][MAX], int n)
{
cout << "Principal Diagonal: ";
for (int i = 0; i < n; i++) {
// Printing principal diagonal
cout << mat[i][i] << ", ";
}
cout << endl;
}
// Function to print the Secondary Diagonal
void printSecondaryDiagonal(int mat[][MAX], int n)
{
cout << "Secondary Diagonal: ";
int k = n - 1;
for (int i = 0; i < n; i++) {
// Printing secondary diagonal
cout << mat[i][k--] << ", ";
}
cout << endl;
}
// Driver code
int main()
{
int n = 4;
int a[][MAX] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
return 0;
}
// This code is contributed by yashbeersingh42
Java
// Java Program to print the
// Diagonals of a Matrix
class Main{
static int MAX = 100;
// Function to print the Principal Diagonal
public static void printPrincipalDiagonal(int mat[][],
int n)
{
System.out.print("Principal Diagonal: ");
for (int i = 0; i < n; i++)
{
// Printing principal diagonal
System.out.print(mat[i][i] + ", ");
}
System.out.println();
}
// Function to print the Secondary Diagonal
public static void printSecondaryDiagonal(int mat[][],
int n)
{
System.out.print("Secondary Diagonal: ");
int k = n - 1;
for (int i = 0; i < n; i++)
{
// Printing secondary diagonal
System.out.print(mat[i][k--] + ", ");
}
System.out.println();
}
public static void main(String[] args)
{
int n = 4;
int a[][] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{1, 2, 3, 4},
{5, 6, 7, 8}};
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program to print the
# Diagonals of a Matrix
MAX = 100
# Function to print the Principal Diagonal
def printPrincipalDiagonal(mat, n):
print("Principal Diagonal: ", end = "")
for i in range(n):
# Printing principal diagonal
print(mat[i][i], end = ", ")
print()
# Function to print the Secondary Diagonal
def printSecondaryDiagonal(mat, n):
print("Secondary Diagonal: ", end = "")
k = n - 1
for i in range(n):
# Printing secondary diagonal
print(mat[i][k], end = ", ")
k -= 1
print()
# Driver Code
n = 4
a = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ] ]
printPrincipalDiagonal(a, n)
printSecondaryDiagonal(a, n)
# This code is contributed by divyesh072019
C#
// C# program for the
// above approach
using System;
class GFG{
// Function to print the
// Principal Diagonal
static void printPrincipalDiagonal(int [,]mat,
int n)
{
Console.Write("Principal Diagonal: ");
for (int i = 0; i < n; i++)
{
// Printing principal diagonal
Console.Write(mat[i, i] + ", ");
}
Console.Write("\n");
}
// Function to print the
// Secondary Diagonal
static void printSecondaryDiagonal(int [,]mat,
int n)
{
Console.Write("Secondary Diagonal: ");
int k = n - 1;
for (int i = 0; i < n; i++)
{
// Printing secondary diagonal
Console.Write(mat[i, k--] + ", ");
}
Console.Write("\n");
}
// Driver code
static void Main()
{
int n = 4;
int [,]a = {{1, 2, 3, 4},
{5, 6, 7, 8},
{1, 2, 3, 4},
{5, 6, 7, 8}};
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
}
}
// This code is contributed by rutvik_56
Javascript
<script>
// Javascript Program to print the
// Diagonals of a Matrix
let MAX = 100;
// Function to print the Principal Diagonal
function printPrincipalDiagonal(mat, n)
{
document.write("Principal Diagonal: ");
for (let i = 0; i < n; i++)
{
// Printing principal diagonal
document.write(mat[i][i] + ", ");
}
document.write("</br>");
}
// Function to print the Secondary Diagonal
function printSecondaryDiagonal(mat, n)
{
document.write("Secondary Diagonal: ");
let k = n - 1;
for (let i = 0; i < n; i++)
{
// Printing secondary diagonal
document.write(mat[i][k--] + ", ");
}
document.write("</br>");
}
let n = 4;
let a = [[1, 2, 3, 4],
[5, 6, 7, 8],
[1, 2, 3, 4],
[5, 6, 7, 8]];
printPrincipalDiagonal(a, n);
printSecondaryDiagonal(a, n);
</script>
Producción:
Principal Diagonal: 1, 6, 3, 8, Secondary Diagonal: 4, 7, 2, 5,
Análisis de Complejidad:
- Complejidad temporal: O(n).
Como solo hay un bucle involucrado, la complejidad del tiempo es lineal. - Espacio Auxiliar: O(1).
Como no se ocupa espacio extra.
Publicación traducida automáticamente
Artículo escrito por sunilkannur98 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA