Element in a matrix starting from which anti-clockwise traversal ends at the last element

Given a mat[][] of size n X n, the task is to find an element X such that if the anti-clockwise traversal is begun from X then the final element to be printed is mat[n – 1][n – 1].

The anti-clockwise traversal of the matrix, mat[][] =
{{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
starting at element 5 will be 5, 6, 3, 2, 1, 4, 7, 8, 9.

Examples:

Input: mat[][] = {{1, 2}, {3, 4}}
Output: 2
If we start traversing from mat[0][1] i.e. 2 then
we will end up with the element at mat[1][1] which is 4.

Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output: 5



Approach: Starting from the element at mat[n – 1][n – 1], start traversing the matrix in the opposite order i.e. clockwise. When all the elements of the matrix are traversed, the last visited element will be the result.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to print last element
// Of given matrix
int printLastElement(int mat[][2], int n)
{
  
    // Starting row index
    int si = 0;
  
    // Starting column index
    int sj = 0;
  
    // Ending row index
    int ei = n - 1;
  
    // Ending column index
    int ej = n - 1;
  
    // Track the move
    int direction = 0;
  
    // While starting index is less than ending
    // row index or starting column index
    // is less the ending column index
    while (si < ei || sj < ej) {
  
        // Switch cases for all direction
        // Move under all cases for all
        // directions
        switch (direction % 4) {
        case 0:
            sj++;
            break;
        case 1:
            ei--;
            break;
        case 2:
            ej--;
            break;
        case 3:
            si++;
            break;
        }
  
        // Increament direction by one
        // for each case
        direction++;
    }
  
    // Finally return the last element
    // If not found return 0
    return mat[si][sj];
  
    return 0;
}
  
// Driver code
int main()
{
    int n = 2;
    int mat[][2] = { { 1, 2 }, { 3, 4 } };
    cout << printLastElement(mat, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GfG
  
// Function to print last element 
// Of given matrix 
static int printLastElement(int mat[][], int n) 
  
    // Starting row index 
    int si = 0
  
    // Starting column index 
    int sj = 0
  
    // Ending row index 
    int ei = n - 1
  
    // Ending column index 
    int ej = n - 1
  
    // Track the move 
    int direction = 0
  
    // While starting index is less than ending 
    // row index or starting column index 
    // is less the ending column index 
    while (si < ei || sj < ej) 
    
  
        // Switch cases for all direction 
        // Move under all cases for all 
        // directions 
        switch (direction % 4
        
        case 0
            sj++; 
            break
        case 1
            ei--; 
            break
        case 2
            ej--; 
            break
        case 3
            si++; 
            break
        
  
        // Increament direction by one 
        // for each case 
        direction++; 
    
  
    // Finally return the last element 
    // If not found return 0 
    return mat[si][sj]; 
  
  
// Driver code 
public static void main(String[] args) 
    int n = 2
    int mat[][] = new int[][]{{ 1, 2 }, { 3, 4 }}; 
    System.out.println(printLastElement(mat, n)); 
}
  
// This code is contributed by Prerna Saini

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Python3

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# Python 3 implementation of the approach
  
# Function to print last element
# Of given matrix
def printLastElement(mat, n):
      
    # Starting row index
    si = 0
  
    # Starting column index
    sj = 0
  
    # Ending row index
    ei = n - 1
  
    # Ending column index
    ej = n - 1
  
    # Track the move
    direction = 0
  
    # While starting index is less than ending
    # row index or starting column index
    # is less the ending column index
    while (si < ei or sj < ej):
          
        # Switch cases for all direction
        # Move under all cases for all
        # directions
        if (direction % 4 == 0):
            sj += 1
        if (direction % 4 == 1):
            ei -= 1
        if (direction % 4 == 2):
            ej -= 1
        if (direction % 4 == 3):
            si += 1
      
        # Increament direction by one
        # for each case
        direction += 1
  
    # Finally return the last element
    # If not found return 0
    return mat[si][sj]
  
    return 0
  
# Driver code
if __name__ == '__main__':
    n = 2
    mat = [[1, 2], [3, 4 ]]
    print(printLastElement(mat, n))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the above approach 
using System;
  
class GFG 
  
// Function to print last element 
// Of given matrix 
static int printLastElement(int [,]mat, int n) 
  
    // Starting row index 
    int si = 0; 
  
    // Starting column index 
    int sj = 0; 
  
    // Ending row index 
    int ei = n - 1; 
  
    // Ending column index 
    int ej = n - 1; 
  
    // Track the move 
    int direction = 0; 
  
    // While starting index is less than ending 
    // row index or starting column index 
    // is less the ending column index 
    while (si < ei || sj < ej) 
    
  
        // Switch cases for all direction 
        // Move under all cases for all 
        // directions 
        switch (direction % 4) 
        
            case 0: 
                sj++; 
                break
            case 1: 
                ei--; 
                break
            case 2: 
                ej--; 
                break
            case 3: 
                si++; 
                break
        
  
        // Increament direction by one 
        // for each case 
        direction++; 
    
  
    // Finally return the last element 
    // If not found return 0 
    return mat[si, sj]; 
  
  
// Driver code 
public static void Main() 
    int n = 2; 
    int [,]mat = {{ 1, 2 }, { 3, 4 }}; 
      
    Console.WriteLine(printLastElement(mat, n)); 
  
// This code is contributed by Ryuga

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PHP

Output:

2


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