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++

// 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;
        }

        // Increment 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;
}

Java

// Java implementation of the approach
import java.io.*;
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;
        }

        // Increment 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

Python3

# 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

        # Increment 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

C#

// 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;
        }

        // Increment 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

PHP

<?php
// PHP implementation of the approach

// Function to print last element
// Of given matrix
function 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 || $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;
        }

        // Increment 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
$n = 2;
$mat = array(array(1, 2),
             array(3, 4));
echo printLastElement($mat, $n);

// This code is contributed by Akanksha Rai
?>

Javascript

<script>

// JavaScript implementation of the approach


    // Function to print last element
    // Of given matrix
    function printLastElement(mat , n) {

        // Starting row index
        var si = 0;

        // Starting column index
        var sj = 0;

        // Ending row index
        var ei = n - 1;

        // Ending column index
        var ej = n - 1;

        // Track the move
        var 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;
            }

            // Increment direction by one
            // for each case
            direction++;
        }

        // Finally return the last element
        // If not found return 0
        return mat[si][sj];

    }

    // Driver code

        var n = 2;
        var mat = [[ 1, 2 ], [ 3, 4 ] ];
        document.write(printLastElement(mat, n));

// This code contributed by gauravrajput1

</script>

Output:

2

Time Complexity: O(N²)
Auxiliary Space: O(1)