Given a 2D array, print it in spiral form. See the following examples.
Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Output: 1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11
Solution:
/* This code is adopted from the solution given #include <stdio.h> #define R 3 #define C 6 void spiralPrint( int m, int n, int a[R][C]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { /* Print the first row from the remaining rows */ for (i = l; i < n; ++i) { printf ( "%d " , a[k][i]); } k++; /* Print the last column from the remaining columns */ for (i = k; i < m; ++i) { printf ( "%d " , a[i][n-1]); } n--; /* Print the last row from the remaining rows */ if ( k < m) { for (i = n-1; i >= l; --i) { printf ( "%d " , a[m-1][i]); } m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m-1; i >= k; --i) { printf ( "%d " , 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} }; spiralPrint(R, C, a); return 0; } /* OUTPUT: 1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11 */ |
Time Complexity: Time complexity of the above solution is O(mn).
Please write comments if you find the above code incorrect, or find other ways to solve the same problem.
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