## print two-dimensional array in spiral order

So I saw this problem in a book today about printing a 2d matrix in spiral order

Here are two solutions to it

### Solution one.

```
def printSpiralTL(m,x1,y1,x2,y2):
for i in range(x1,x2):
print m[y1][i]
for j in range(y1+1,y2+1):
print m[j][x2-1]
if x2-x1 > 0:
printSpiralBL(m, x1, y1 + 1, x2-1, y2)
def printSpiralBL(m,x1,y1,x2,y2):
for i in range(x2-1,x1-1,-1):
print m[y2][i]
for j in range(y2-1,y1-1,-1):
print m[j][x1]
if x2-x1 > 0:
printSpiralTL(m, x1+1, y1, x2, y2-1)
m = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6],
[7, 8, 9, 1]
]
```

Output:

```
1
2
3
4
8
2
6
1
9
8
7
3
9
5
6
7
1
5
4
0
```

Here is another way

### Solution two (print - pop -Transpose).

```
m = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6],
[7, 8, 9, 1]
]
def spiral(arr):
while arr:
for a in arr[0]:
print a
arr = list(reversed(zip(*arr[1:])))
spiral(m)
```

Hope this helps.

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