Dada una array arr[][] de tamaño N * M , la tarea es imprimir los elementos de contorno de la array dada en el sentido de las agujas del reloj.
Ejemplos:
Entrada: arr[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9} }
Salida: 1 2 3 6 9 8 7 4
Explicación:
Elementos de contorno de la array son:
1 2 3
4 5 6
7 8 < strong>9
Por lo tanto, la secuencia de elementos de contorno en el sentido de las agujas del reloj es {1, 2, 3, 6, 9, 8, 7, 4} .
Entrada: arr[][] = {{11, 12, 33}, {64, 57, 61}, {74, 88, 39}}
Salida: 11 12 33 61 39 88 74 64
Enfoque ingenuo: el enfoque más simple para resolver este problema es atravesar la array dada y verificar si el elemento actual es el elemento límite o no . Si se encuentra que es cierto, imprima el elemento.
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)
Enfoque eficiente: para optimizar el enfoque anterior, la idea es recorrer solo la primera y la última fila y la primera y la última columna de la array. Siga los pasos a continuación para resolver el problema:
- Imprime la primera fila de la array.
- Imprime la última columna de la array excepto la primera fila.
- Imprime la última fila de la array excepto la última columna.
- Imprime la primera columna de la array excepto la primera y la última fila.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the boundary elements
// of the matrix in clockwise
void boundaryTraversal(vector<vector<int> > arr, int N,
int M)
{
// Print the first row
for (int i = 0; i < M; i++)
{
cout << arr[0][i] << " ";
}
// Print the last column
// except the first row
for (int i = 1; i < N; i++)
{
cout << arr[i][M - 1] << " ";
}
// Print the last row
// except the last column
if (N > 1)
{
// Print the last row
for (int i = M - 2; i >= 0; i--)
{
cout << arr[N - 1][i] << " ";
}
}
// Print the first column except
// the first and last row
if (M > 1) {
// Print the first column
for (int i = N - 2; i > 0; i--) {
cout << arr[i][0] << " ";
}
}
}
// Driver Code
int main()
{
vector<vector<int> > arr{ { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int N = arr.size();
int M = arr[0].size();
// Function Call
boundaryTraversal(arr, N, M);
return 0;
}
// This code is contributed by Dharanendra L V
Java
// Java program of the above approach
import java.util.*;
class GFG {
// Function to print the boundary elements
// of the matrix in clockwise
public static void boundaryTraversal(
int arr[][], int N, int M)
{
// Print the first row
for (int i = 0; i < M; i++) {
System.out.print(arr[0][i] + " ");
}
// Print the last column
// except the first row
for (int i = 1; i < N; i++) {
System.out.print(arr[i][M - 1] + " ");
}
// Print the last row
// except the last column
if (N > 1) {
// Print the last row
for (int i = M - 2; i >= 0; i--) {
System.out.print(arr[N - 1][i] + " ");
}
}
// Print the first column except
// the first and last row
if (M > 1) {
// Print the first column
for (int i = N - 2; i > 0; i--) {
System.out.print(arr[i][0] + " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[][]
= { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int N = arr.length;
int M = arr[0].length;
// Function Call
boundaryTraversal(arr, N, M);
}
}
Python3
# Python program of the above approach # Function to print the boundary elements # of the matrix in clockwise def boundaryTraversal(arr, N, M): # Print the first row for i in range(M): print(arr[0][i], end = " "); # Print the last column # except the first row for i in range(1, N): print(arr[i][M - 1], end = " "); # Print the last row # except the last column if (N > 1): # Print the last row for i in range(M - 2, -1, -1): print(arr[N - 1][i], end = " "); # Print the first column except # the first and last row if (M > 1): # Print the first column for i in range(N - 2, 0, -1): print(arr[i][0], end = " "); # Driver Code if __name__ == '__main__': arr = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]; N = len(arr); M = len(arr[0]); # Function Call boundaryTraversal(arr, N, M); # This code is contributed by 29AjayKumar
C#
// C# program of the above approach
using System;
class GFG{
// Function to print the boundary elements
// of the matrix in clockwise
static void boundaryTraversal(int[,] arr,
int N, int M)
{
// Print the first row
for(int i = 0; i < M; i++)
{
Console.Write(arr[0, i] + " ");
}
// Print the last column
// except the first row
for(int i = 1; i < N; i++)
{
Console.Write(arr[i, M - 1] + " ");
}
// Print the last row
// except the last column
if (N > 1)
{
// Print the last row
for(int i = M - 2; i >= 0; i--)
{
Console.Write(arr[N - 1, i] + " ");
}
}
// Print the first column except
// the first and last row
if (M > 1)
{
// Print the first column
for(int i = N - 2; i > 0; i--)
{
Console.Write(arr[i, 0] + " ");
}
}
}
// Driver code
static void Main()
{
int[,] arr = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int N = 3;
int M = 3;
// Function Call
boundaryTraversal(arr, N, M);
}
}
// This code is contributed by divyeshrabadiya07
Javascript
<script>
// Javascript program of the above approach
// Function to print the boundary elements
// of the matrix in clockwise
function boundaryTraversal(arr, N, M)
{
// Print the first row
for (let i = 0; i < M; i++)
{
document.write(arr[0][i] + " ");
}
// Print the last column
// except the first row
for (let i = 1; i < N; i++)
{
document.write(arr[i][M - 1] + " ");
}
// Print the last row
// except the last column
if (N > 1)
{
// Print the last row
for (let i = M - 2; i >= 0; i--)
{
document.write(arr[N - 1][i] + " ");
}
}
// Print the first column except
// the first and last row
if (M > 1) {
// Print the first column
for (let i = N - 2; i > 0; i--) {
document.write(arr[i][0] + " ");
}
}
}
// Driver Code
let arr = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ];
let N = arr.length;
let M = arr[0].length;
// Function Call
boundaryTraversal(arr, N, M);
</script>
Producción:
1 2 3 6 9 8 7 4
Complejidad temporal: O(N + M)
Espacio auxiliar: O(1)