Print boundary elements of a given matrix in clockwise manner

0 0 5 minutes read

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 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>

Output:

1 2 3 6 9 8 7 4

Time Complexity: O(N + M)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!