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4700297b3e
* Enable ruff RUF002 rule * Fix --------- Co-authored-by: Christian Clauss <cclauss@me.com>
134 lines
3.5 KiB
Python
134 lines
3.5 KiB
Python
"""
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This program print the matrix in spiral form.
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This problem has been solved through recursive way.
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Matrix must satisfy below conditions
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i) matrix should be only one or two dimensional
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ii) number of column of all rows should be equal
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"""
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def check_matrix(matrix: list[list[int]]) -> bool:
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# must be
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matrix = [list(row) for row in matrix]
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if matrix and isinstance(matrix, list):
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if isinstance(matrix[0], list):
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prev_len = 0
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for row in matrix:
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if prev_len == 0:
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prev_len = len(row)
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result = True
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else:
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result = prev_len == len(row)
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else:
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result = True
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else:
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result = False
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return result
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def spiral_print_clockwise(a: list[list[int]]) -> None:
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"""
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>>> spiral_print_clockwise([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
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1
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2
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3
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4
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8
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12
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11
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10
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9
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5
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6
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7
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"""
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if check_matrix(a) and len(a) > 0:
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a = [list(row) for row in a]
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mat_row = len(a)
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if isinstance(a[0], list):
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mat_col = len(a[0])
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else:
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for dat in a:
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print(dat)
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return
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# horizotal printing increasing
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for i in range(mat_col):
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print(a[0][i])
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# vertical printing down
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for i in range(1, mat_row):
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print(a[i][mat_col - 1])
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# horizotal printing decreasing
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if mat_row > 1:
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for i in range(mat_col - 2, -1, -1):
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print(a[mat_row - 1][i])
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# vertical printing up
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for i in range(mat_row - 2, 0, -1):
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print(a[i][0])
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remain_mat = [row[1 : mat_col - 1] for row in a[1 : mat_row - 1]]
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if len(remain_mat) > 0:
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spiral_print_clockwise(remain_mat)
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else:
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return
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else:
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print("Not a valid matrix")
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return
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# Other Easy to understand Approach
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def spiral_traversal(matrix: list[list]) -> list[int]:
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"""
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>>> spiral_traversal([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
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[1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]
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Example:
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matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
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Algorithm:
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Step 1. first pop the 0 index list. (which is [1,2,3,4] and concatenate the
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output of [step 2])
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Step 2. Now perform matrix's Transpose operation (Change rows to column
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and vice versa) and reverse the resultant matrix.
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Step 3. Pass the output of [2nd step], to same recursive function till
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base case hits.
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Dry Run:
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Stage 1.
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[1, 2, 3, 4] + spiral_traversal([
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[8, 12], [7, 11], [6, 10], [5, 9]]
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])
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Stage 2.
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[1, 2, 3, 4, 8, 12] + spiral_traversal([
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[11, 10, 9], [7, 6, 5]
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])
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Stage 3.
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[1, 2, 3, 4, 8, 12, 11, 10, 9] + spiral_traversal([
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[5], [6], [7]
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])
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Stage 4.
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[1, 2, 3, 4, 8, 12, 11, 10, 9, 5] + spiral_traversal([
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[5], [6], [7]
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])
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Stage 5.
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[1, 2, 3, 4, 8, 12, 11, 10, 9, 5] + spiral_traversal([[6, 7]])
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Stage 6.
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[1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7] + spiral_traversal([])
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"""
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if matrix:
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return list(matrix.pop(0)) + spiral_traversal(
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[list(row) for row in zip(*matrix)][::-1]
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)
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else:
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return []
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# driver code
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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a = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
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spiral_print_clockwise(a)
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