Print K’th element in spiral form of matrix

Given a 2D Matrix of order n X m , print K’th element in spiral form of 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

Input: mat[][] =
{{1, 2, 3, 4, 5, 6}
{7, 8, 9, 10, 11, 12}
{13, 14, 15, 16, 17, 18}}
k = 17
Output: 10

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

Simple Solution:
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. Time complexity for this approach will be O(n^2).

Efficient Solution:
We will consider only edge of the matrix at a time and then do recursion for the sub matrix made by removing edges of main matrix.

1. If k <= m : element is in first row.
2. Else If k <= (m+n-1) : element is in last column.
3. Else If k <= (m+n-1+m-1) : element is in last row.
4. Else If k <= (m+n-1+m-1+n-2) : element is in first column.
5. Else Element lies somewhere in middle matrix.

 // 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 program to test above functions */ 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; }

Output:

10

Time Complexity : O(c) where c is number of outer circular rings with respect to k’th element.
Space complexity: O(1)

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