# 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 i.e. 2 then
we will end up with the element at mat which is 4.

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Function to print last element ` `// Of given matrix ` `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]; ` ` `  `    ``return` `0; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 2; ` `    ``int` `mat[] = { { 1, 2 }, { 3, 4 } }; ` `    ``cout << printLastElement(mat, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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``;  ` `        ``}  ` ` `  `        ``// 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

 ` `

Output:

```2
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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