Dado un número entero N y una array N x N , la tarea es convertir la array dada en una array simétrica reemplazando (i, j) th y (j, i) th elemento con su media aritmética .
Ejemplos:
Entrada: arr[] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
Salida:
1 3 5
3 5 7
5 7 9
Explicación: Los elementos de la diagonal son iguales. El elemento en el índice (0, 1) = 2 y (1, 0) = 4 se reemplaza por su media aritmética, es decir, (2 + 4) / 2 = 3. De manera similar, los elementos en el índice (2, 0) y ( 0, 2), (2, 1) y (1, 2) también se reemplazan por su media aritmética y la array de salida resultante es una array simétrica.Entrada: arr[] = {{12, 43, 65},
{23, 75, 13},
{51, 37, 81}}
Salida:
12 33 58
33 75 25
58 25 81
Enfoque: El problema dado es un problema basado en la implementación. La idea es recorrer la array triangular inferior y reemplazar los elementos y sus respectivos índices de transposición con su media aritmética .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program of the above approach
#include <iostream>
using namespace std;
const int N = 3;
// Function to convert the given matrix
// into a symmetric matrix by replacing
// transpose elements with their mean
void makeSymmetric(int mat[][N])
{
// Loop to traverse lower triangular
// elements of the given matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (j < i) {
mat[i][j] = mat[j][i]
= (mat[i][j] +
mat[j][i]) / 2;
}
}
}
}
// Function to print the given matrix
void showMatrix(int mat[][N])
{
// Loop to traverse the
// given matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
// Print current index
cout << mat[i][j] << " ";
}
cout << "\n";
}
}
// Driver Code
int main()
{
int arr[][N]
= { { 12, 43, 65 },
{ 23, 75, 13 },
{ 51, 37, 81 } };
makeSymmetric(arr);
showMatrix(arr);
return 0;
}
Java
// Java program of the above approach
import java.util.*;
class GFG{
static int N = 3;
// Function to convert the given matrix
// into a symmetric matrix by replacing
// transpose elements with their mean
static void makeSymmetric(int mat[][])
{
// Loop to traverse lower triangular
// elements of the given matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if (j < i)
{
mat[i][j] = mat[j][i] = (mat[i][j] +
mat[j][i]) / 2;
}
}
}
}
// Function to print the given matrix
static void showMatrix(int mat[][])
{
// Loop to traverse the
// given matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// Print current index
System.out.print(mat[i][j] + " ");
}
System.out.println();
}
}
// Driver Code
public static void main(String args[])
{
int arr[][] = { { 12, 43, 65 },
{ 23, 75, 13 },
{ 51, 37, 81 } };
makeSymmetric(arr);
showMatrix(arr);
}
}
// This code is contributed by sanjoy_62
Python3
# python3 program of the above approach N = 3 # Function to convert the given matrix # into a symmetric matrix by replacing # transpose elements with their mean def makeSymmetric(mat): # Loop to traverse lower triangular # elements of the given matrix for i in range(0, N): for j in range(0, N): if (j < i): mat[i][j] = mat[j][i] = (mat[i][j] + mat[j][i]) // 2 # Function to print the given matrix def showMatrix(mat): # Loop to traverse the # given matrix for i in range(0, N): for j in range(0, N): # Print current index print(mat[i][j], end=" ") print() # Driver Code if __name__ == "__main__": arr = [[12, 43, 65], [23, 75, 13], [51, 37, 81]] makeSymmetric(arr) showMatrix(arr) # This code is contributed by rakeshsahni
C#
// C# program of the above approach
using System;
public class GFG
{
static int N = 3;
// Function to convert the given matrix
// into a symmetric matrix by replacing
// transpose elements with their mean
static void makeSymmetric(int [,]mat)
{
// Loop to traverse lower triangular
// elements of the given matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if (j < i)
{
mat[i,j] = mat[j,i] = (mat[i,j] +
mat[j,i]) / 2;
}
}
}
}
// Function to print the given matrix
static void showMatrix(int [,]mat)
{
// Loop to traverse the
// given matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// Print current index
Console.Write(mat[i, j] + " ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main(String []args)
{
int [,]arr = { { 12, 43, 65 },
{ 23, 75, 13 },
{ 51, 37, 81 } };
makeSymmetric(arr);
showMatrix(arr);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript code for the above approach
let N = 3;
// Function to convert the given matrix
// into a symmetric matrix by replacing
// transpose elements with their mean
function makeSymmetric(mat)
{
// Loop to traverse lower triangular
// elements of the given matrix
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
if (j < i) {
mat[i][j] = mat[j][i]
= Math.floor((mat[i][j] +
mat[j][i]) / 2);
}
}
}
}
// Function to print the given matrix
function showMatrix(mat)
{
// Loop to traverse the
// given matrix
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
// Print current index
document.write(mat[i][j] + " ");
}
document.write('<br>')
}
}
// Driver Code
let arr = [[12, 43, 65],
[23, 75, 13],
[51, 37, 81]];
makeSymmetric(arr);
showMatrix(arr);
// This code is contributed by Potta Lokesh
</script>
12 33 58 33 75 25 58 25 81
Complejidad temporal: O(N 2 )
Complejidad espacial: O(1)
Publicación traducida automáticamente
Artículo escrito por pintusaini y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA