Given a 2D array, the task is to print matrix in anti spiral form:
Examples:
Output: 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Input : arr[][4] = {1, 2, 3, 4
5, 6, 7, 8
9, 10, 11, 12
13, 14, 15, 16};
Output : 10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1
Input :arr[][6] = {1, 2, 3, 4, 5, 6
7, 8, 9, 10, 11, 12
13, 14, 15, 16, 17, 18};
Output : 11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1
The idea is simple, we traverse matrix in spiral form and put all traversed elements in a stack. Finally one by one elements from stack and print them.
Javascript
<script>// Javascript Code for Print matrix in antispiral form function antiSpiralTraversal(m,n,a) { let i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ let stk=[]; while (k <= m && l <= n) { /* Print the first row from the remaining rows */ for (i = l; i <= n; ++i) stk.push(a[k][i]); k++; /* Print the last column from the remaining columns */ for (i = k; i <= m; ++i) stk.push(a[i][n]); n--; /* Print the last row from the remaining rows */ if ( k <= m) { for (i = n; i >= l; --i) stk.push(a[m][i]); m--; } /* Print the first column from the remaining columns */ if (l <= n) { for (i = m; i >= k; --i) stk.push(a[i][l]); l++; } } while (stk.length!=0) { document.write(stk[stk.length-1] + " "); stk.pop(); } } /* Driver program to test above function */ let mat = [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20]]; antiSpiralTraversal(mat.length - 1, mat[0].length - 1, mat); // This code is contributed by avanitrachhadiya2155</script> |
Output:
12 13 14 9 8 7 6 11 16 17 18 19 20 15 10 5 4 3 2 1
Time complexity: O(m*n) where m is no of rows and n is no of columns of a given matrix
Auxiliary Space: O(n)
Please refer complete article on Print matrix in antispiral form for more details!
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