Anticlockwise spiral view of diamond matrix elements in python

Python program for Anticlockwise spiral view of diamond matrix elements. Here mentioned other language solution.

``````#  Python 3 program for
#  Anticlockwise spiral view of diamond element in matrix
class SpiralTraversal :
#  Method which is displaying the Anticlockwise spiral
#  Of matrix in diamond(mid) element
def spiralView(self, matrix,  x,  y,  p,  q,  k) :
#  Find the middle column
midCol = y + (int((q - y) / 2))
#  Get middle row (valid for odd square matrix)
midRow = midCol
#  Matrix are divided into 4 section
#  Starting of middle row and column in form of top
#  Case A
#  Top to Left-bottom
i = midCol
j = 0
while (i > y and k > 0) :
print(matrix[x + j][i], end =" ")
i -= 1
k -= 1
j += 1
#  Case B
#  Middle left to middle right bottom
i = y
j = 0
while (i <= midCol and k > 0) :
print(matrix[(midRow) + j][i], end =" ")
i += 1
k -= 1
j += 1
#  Case C
#  Middle bottom to middle right side
i = midCol + 1
j = 1
while (i <= q and k > 0) :
print(matrix[(p) - j][i], end =" ")
i += 1
k -= 1
j += 1
#  Case D
#  Middle right side to  top middle
i = q - 1
j = 1
while (i > midCol and k > 0) :
print(matrix[(midRow) - j][i], end =" ")
i -= 1
k -= 1
j += 1
if (x + 1 <= p - 1 and k > 0) :
#  Recursive call
self.spiralView(matrix, x + 1, y + 1, p - 1, q - 1, k)
def diamondAnticlockwise(self, matrix) :
row = len(matrix)
col = len(matrix[0])
#  Check whether the given matrix is a valid odd shape or not
if (row != col or col % 2 == 0) :
#  When Yes not a valid size
print("\nNot a valid perfect Odd square matrix")
return
#  (row*col)-((col+1)/2)*4 are provide the value of number of
#  Diamond element
self.spiralView(matrix, 0, 0, row - 1,
col - 1, (row * col) - (int((col + 1) / 2)) * 4)
def main() :
#  Define matrix
matrix = [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, -1, 15, 16],
[17, 18, 19, 20, 21],
[22, 23, 24, 25, 26]
]

if __name__=="__main__":
main()``````

Output

``3 7 11 18 24 20 16 9 8 12 19 15 -1``

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