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Sorting boundary elements of a matrix
  • Difficulty Level : Expert
  • Last Updated : 10 Jun, 2021

Given a matrix mat[][] of size M*N, the task is to sort only the border elements of the matrix in the clockwise direction and print the matrix after sorting again.

Examples: 
 

Input: M = 4, N = 5, Below is the given matrix: 

1 2 3 4 0 
1 1 1 1 2  
1 2 2 2 4 
1 9 3 1 7

Output: 
0 1 1 1 1 
9 1 1 1 1 
7 2 2 2 2 
4 4 3 3 2 
Explanation: 
For given matrix, border elements are: 
(1, 2, 3, 4, 0, 2, 4, 7, 1, 3, 9, 1, 1, 1) 
After sorting in clockwise order: 
(0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 7, 9)

Input: M = 3, N = 4 



4 2 8 0 
2 6 9 8 
0 3 1 7

Output: 
0 0 1 2 
8 6 9 2 
8 7 4 3 
Explanation: 
For given matrix, border elements are: 
(4, 2, 8, 0, 8, 7, 1, 3, 0, 2) 
After sorting in clockwise order: 
(0, 0, 1, 2, 2, 3, 4, 7, 8, 8) 
 

Approach: The idea is to store all border elements of the given matrix in an array and sort this array then simply print the new matrix using this sorted array as the border elements.

Detailed steps are as follows:  

  • Traverse the given matrix and push all the boundary elements to an array A[].
  • Sort array A[] in ascending order.
  • Print first row using first N elements of array A[].
  • From second row to second-last row, first print a single element from end of A[], then print N-2 middle elements from original matrix and finally a single element from the front of A[].
  • For last row, print middle elements from A[] which are still not printed, in reverse order.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
void printMatrix(int grid[][5], int m, int n)
{
  vector<int> A;
 
  // Appending border elements
  for (int i = 0; i < m; i++)
  {
    for (int j = 0; j < n; j++)
    {
      if (j == n - 1 || (i == m - 1) || j == 0
          || i == 0)
        A.push_back(grid[i][j]);
    }
  }
 
  // Sorting the list
  sort(A.begin(), A.end());
 
  // Printing first row with
  // first N elements from A
  for (int i = 0; i < n; i++)
    cout << A[i] << " ";
  cout << endl;
  // print(*A[:n])
 
  // Printing N-2 rows
  for (int i = 0; i < m - 2; i++)
  {
 
    // Print elements from last
    cout << A[A.size() - i - 1] << " ";
 
    // Print middle elements
    // from original matrix
    for (int j = 1; j < n - 1; j++)
      cout << grid[i + 1][j] << " ";
 
    // Print elements from front
    cout << (A[n + i]) << endl;
  }
 
  // Printing last row
  reverse(A.begin() + n + m - 2,
          A.begin() + n + m + n - 2);
  for (int i = n + m - 2; i < n + m - 2 + n; i++)
    cout << A[i] << " ";
  //[n + m - 2:n + m - 2 + n] << endl;
}
 
// Driver Code
int main()
{
  // Dimensions of a Matrix
  int m = 4, n = 5;
 
  // Given Matrix
  int grid[][5] = { { 1, 2, 3, 4, 0 },
                   { 1, 1, 1, 1, 2 },
                   { 1, 2, 2, 2, 4 },
                   { 1, 9, 3, 1, 7 } };
 
  // Function Call
  printMatrix(grid, m, n);
  return 0;
}
 
// This code is contributed by chitranayal.

Python3




# Python program for the above approach
 
def printMatrix(grid, m, n):
     
    A =[]
 
    # Appending border elements
    for i in range(m):
        for j in range(n):
            if j == n-1 or (i == m-1
                ) or j == 0 or i == 0:
                A.append(grid[i][j])
                 
    # Sorting the list
     
    A.sort()
 
 
    # Printing first row with
    # first N elements from A
    print(*A[:n])
 
 
    # Printing N-2 rows
    for i in range(m-2):
         
        # Print elements from last
        print(A[len(A)-i-1],
                          end =" ")
        # Print middle elements
        # from original matrix
        for j in range(1, n-1):
            print(grid[i + 1][j],
                          end =" ")
             
             
        # Print elements from front
        print(A[n + i])
 
    # Printing last row
    print(*reversed(A[n + m-2:n + m-2 + n]))
 
# Driver Code
 
# Dimensions of a Matrix
m, n = 4, 5
 
# Given Matrix
grid =[[1, 2, 3, 4, 0],
       [1, 1, 1, 1, 2],
       [1, 2, 2, 2, 4],
       [1, 9, 3, 1, 7]]
 
# Function Call
printMatrix(grid, m, n)

Javascript




<script>
 
// Javascript program for the above approach
function printMatrix(grid, m, n)
{
    let A = [];
     
    // Appending border elements
    for(let i = 0; i < m; i++)
    {
        for(let j = 0; j < n; j++)
        {
            if (j == n - 1 || (i == m - 1) ||
                j == 0 || i == 0)
                A.push(grid[i][j]);
        }
    }
     
    // Sorting the list
    A.sort(function(a, b){return a - b;});
     
    // Printing first row with
    // first N elements from A
    for(let i = 0; i < n; i++)
        document.write(A[i] + " ");
         
    document.write("<br>")
    // print(*A[:n])
     
    // Printing N-2 rows
    for(let i = 0; i < m - 2; i++)
    {
     
        // Print elements from last
        document.write(A[A.length - i - 1] + " ");
         
        // Print middle elements
        // from original matrix
        for(let j = 1; j < n - 1; j++)
            document.write(grid[i + 1][j] + " ");
         
        // Print elements from front
        document.write(A[n + i] + "<br>")
    }
     
    // Printing last row
    document.write(A.slice(
        n + m - 2, n + m - 2 + n).reverse().join(" "));
    //[n + m - 2:n + m - 2 + n] << endl;
}
 
// Driver Code
 
// Dimensions of a Matrix
let m = 4, n = 5;
 
// Given Matrix
let grid = [ [ 1, 2, 3, 4, 0 ],
             [ 1, 1, 1, 1, 2 ],
             [ 1, 2, 2, 2, 4 ],
             [ 1, 9, 3, 1, 7 ] ];
                    
// Function Call
printMatrix(grid, m, n);
   
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
0 1 1 1 1
9 1 1 1 1
7 2 2 2 2
4 4 3 3 2

 

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

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