Given a matrix arr[][] of size N * M, the task is to print the boundary elements of the given matrix in a clockwise form.
Examples:
Input: arr[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9} }
Output: 1 2 3 6 9 8 7 4
Explanation:
Boundary elements of the matrix are:
1 2 3
4 5 6
7 8 <strong>9
Therefore, the sequence of boundary elements in clockwise form is {1, 2, 3, 6, 9, 8, 7, 4}.
Input: arr[][] = {{11, 12, 33}, {64, 57, 61}, {74, 88, 39}}
Output: 11 12 33 61 39 88 74 64
Naive Approach: The simplest approach to solve this problem is to traverse the given matrix and check if the current element is the boundary element or not. If found to be true, then print the element.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to traverse only the first and last rows and the first and last columns of the matrix. Follow the steps below to solve the problem:
- Print the first row of the matrix.
- Print the last column of the matrix except the first row.
- Print the last row of the matrix except the last column.
- Print the first column of the matrix except the first and last row.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// Function to print the boundary elements// of the matrix in clockwisevoid 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 Codeint 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 approachimport 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 clockwisedef 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 Codeif __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 approachusing 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 clockwisefunction 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> |
Output:
1 2 3 6 9 8 7 4
Time Complexity: O(N + M)
Auxiliary Space: O(1)
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