# Print K’th element in spiral form of matrix

Given a 2D Matrix of order n X m, print K’th element in the spiral form of the matrix. See the following examples.
Examples:

```Input: mat[][] =
{{1, 2, 3, 4}
{5, 6, 7, 8}
{9, 10, 11, 12}
{13, 14, 15, 16}}
k = 6
Output: 12
Explanation: The elements in spiral order is
1, 2, 3, 4, 8, 12, 16, 15...
so the 6th element is 12

Input: mat[][] =
{{1, 2, 3, 4, 5, 6}
{7, 8, 9, 10, 11, 12}
{13, 14, 15, 16, 17, 18}}
k = 17
Output: 10
Explanation: The elements in spiral order is
1, 2, 3, 4, 5, 6, 12, 18, 17,
16, 15, 14, 13, 7, 8, 9, 10, 11
so the 17 th element is 10.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Simple Approah: One simple solution is to start traversing matrix in spiral form Print Spiral Matrix and start a counter i.e; count = 0. Whenever count gets equal to K, print that element.

• Algorithm:
1. Keep a variable count = 0 to store the count.
2. Traverse the matrix in sprial from start to end.
3. Increase the count by 1 for every iteration.
4. If the count is equal to the given value of k print the element and break.
• Implementation

 `#include ` `using` `namespace` `std; ` `#define R 3 ` `#define C 6 ` ` `  `void` `spiralPrint(``int` `m, ``int` `n, ``int` `a[R][C], ``int` `c) ` `{ ` `    ``int` `i, k = 0, l = 0; ` `    ``int` `count = 0; ` ` `  `    ``/* k - starting row index  ` `        ``m - ending row index  ` `        ``l - starting column index  ` `        ``n - ending column index  ` `        ``i - iterator  ` `    ``*/` ` `  `    ``while` `(k < m && l < n) { ` `        ``/* check the first row from  ` `            ``the remaining rows */` `        ``for` `(i = l; i < n; ++i) { ` `            ``count++; ` ` `  `            ``if` `(count == c) ` `                ``cout << a[k][i] << ``" "``; ` `        ``} ` `        ``k++; ` ` `  `        ``/* check the last column  ` `        ``from the remaining columns */` `        ``for` `(i = k; i < m; ++i) { ` `            ``count++; ` ` `  `            ``if` `(count == c) ` `                ``cout << a[i][n - 1] << ``" "``; ` `        ``} ` `        ``n--; ` ` `  `        ``/* check the last row from  ` `                ``the remaining rows */` `        ``if` `(k < m) { ` `            ``for` `(i = n - 1; i >= l; --i) { ` `                ``count++; ` ` `  `                ``if` `(count == c) ` `                    ``cout << a[m - 1][i] << ``" "``; ` `            ``} ` `            ``m--; ` `        ``} ` ` `  `        ``/* check the first column from  ` `                ``the remaining columns */` `        ``if` `(l < n) { ` `            ``for` `(i = m - 1; i >= k; --i) { ` `                ``count++; ` ` `  `                ``if` `(count == c) ` `                    ``cout << a[i][l] << ``" "``; ` `            ``} ` `            ``l++; ` `        ``} ` `    ``} ` `} ` ` `  `/* Driver program to test above functions */` `int` `main() ` `{ ` `    ``int` `a[R][C] = { { 1, 2, 3, 4, 5, 6 }, ` `                    ``{ 7, 8, 9, 10, 11, 12 }, ` `                    ``{ 13, 14, 15, 16, 17, 18 } }, ` `        ``k = 17; ` ` `  `    ``spiralPrint(R, C, a, k); ` `    ``return` `0; ` `} `

Output:

```10
```
• Complexity Analysis:
• Time Complexity : O(R*C), A single traversal of matrix is needed so the Time Complexity is O(R*C).
• Space Complexity : O(1), constant space is required.

Efficient Approach: While traversing the array in spiral order, a loop is used to traverse the sides. So if it can be found out that the kth element is in the given side then the kth element can be found out in constant time. This can be done recursively as well as iteratively.

• Algorithm :
1. Traverse the matrix in form of spiral or cycles.
2. So a cycle can be divided into 4 parts, so if the cycle is of size m X n.
3. Element is in first row, i.e k <= m
4. Element is in last column, i.e k <= (m+n-1)
5. Element is in last row, i.e. k <= (m+n-1+m-1)
6. Element is in first column, i.e k <= (m+n-1+m-1+n-2)
7. If any of the above conditions meet then the kth element can be found is constant time.
8. Else remove the cycle from the array and recursively call the function.
• Implementation:

## C++

 `// C++ program for Kth element in spiral ` `// form of matrix ` `#include ` `#define MAX 100 ` `using` `namespace` `std; ` ` `  `/* function for Kth element */` `int` `findK(``int` `A[MAX][MAX], ``int` `n, ``int` `m, ``int` `k) ` `{ ` `    ``if` `(n < 1 || m < 1) ` `        ``return` `-1; ` ` `  `    ``/*....If element is in outermost ring ....*/` `    ``/* Element is in first row */` `    ``if` `(k <= m) ` `        ``return` `A[k - 1]; ` ` `  `    ``/* Element is in last column */` `    ``if` `(k <= (m + n - 1)) ` `        ``return` `A[(k - m)][m - 1]; ` ` `  `    ``/* Element is in last row */` `    ``if` `(k <= (m + n - 1 + m - 1)) ` `        ``return` `A[n - 1][m - 1 - (k - (m + n - 1))]; ` ` `  `    ``/* Element is in first column */` `    ``if` `(k <= (m + n - 1 + m - 1 + n - 2)) ` `        ``return` `A[n - 1 - (k - (m + n - 1 + m - 1))]; ` ` `  `    ``/*....If element is NOT in outermost ring ....*/` `    ``/* Recursion for sub-matrix. &A is ` `    ``address to next inside sub matrix.*/` `    ``return` `findK((``int``(*)[MAX])(&(A)), n - 2, ` `                 ``m - 2, k - (2 * n + 2 * m - 4)); ` `} ` ` `  `/* Driver code */` `int` `main() ` `{ ` `    ``int` `a[MAX][MAX] = { { 1, 2, 3, 4, 5, 6 }, ` `                        ``{ 7, 8, 9, 10, 11, 12 }, ` `                        ``{ 13, 14, 15, 16, 17, 18 } }; ` `    ``int` `k = 17; ` `    ``cout << findK(a, 3, 6, k) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program for Kth element in spiral ` `// form of matrix ` `class` `GFG { ` ` `  `    ``static` `int` `MAX = ``100``; ` ` `  `    ``/* function for Kth element */` `    ``static` `int` `findK(``int` `A[][], ``int` `i, ``int` `j, ` `                     ``int` `n, ``int` `m, ``int` `k) ` `    ``{ ` `        ``if` `(n < ``1` `|| m < ``1``) ` `            ``return` `-``1``; ` ` `  `        ``/*.....If element is in outermost ring ....*/` `        ``/* Element is in first row */` `        ``if` `(k <= m) ` `            ``return` `A[i + ``0``][j + k - ``1``]; ` ` `  `        ``/* Element is in last column */` `        ``if` `(k <= (m + n - ``1``)) ` `            ``return` `A[i + (k - m)][j + m - ``1``]; ` ` `  `        ``/* Element is in last row */` `        ``if` `(k <= (m + n - ``1` `+ m - ``1``)) ` `            ``return` `A[i + n - ``1``][j + m - ``1` `- (k - (m + n - ``1``))]; ` ` `  `        ``/* Element is in first column */` `        ``if` `(k <= (m + n - ``1` `+ m - ``1` `+ n - ``2``)) ` `            ``return` `A[i + n - ``1` `- (k - (m + n - ``1` `+ m - ``1``))][j + ``0``]; ` ` `  `        ``/*.....If element is NOT in outermost ring ....*/` `        ``/* Recursion for sub-matrix. &A is ` `    ``address to next inside sub matrix.*/` `        ``return` `findK(A, i + ``1``, j + ``1``, n - ``2``, ` `                     ``m - ``2``, k - (``2` `* n + ``2` `* m - ``4``)); ` `    ``} ` ` `  `    ``/* Driver code */` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `a[][] = { { ``1``, ``2``, ``3``, ``4``, ``5``, ``6` `}, ` `                      ``{ ``7``, ``8``, ``9``, ``10``, ``11``, ``12` `}, ` `                      ``{ ``13``, ``14``, ``15``, ``16``, ``17``, ``18` `} }; ` `        ``int` `k = ``17``; ` `        ``System.out.println(findK(a, ``0``, ``0``, ``3``, ``6``, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

Output:

``` 10
```
• Complexity Analysis:
• Time Complexity: O(c), where c is number of outer circular rings with respect to k’th element.
• Space Complexity: O(1).
As constant space is required.

This article is contributed by Shashank Mishra (Gullu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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