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76 lines
2.1 KiB
Python
76 lines
2.1 KiB
Python
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"""
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Given a grid, where you start from the top left position [0, 0],
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you want to find how many paths you can take to get to the bottom right position.
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start here -> 0 0 0 0
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1 1 0 0
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0 0 0 1
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0 1 0 0 <- finish here
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how many 'distinct' paths can you take to get to the finish?
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Using a recursive depth-first search algorithm below, you are able to
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find the number of distinct unique paths (count).
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'*' will demonstrate a path
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In the example above, there are two distinct paths:
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1. 2.
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* * * 0 * * * *
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1 1 * 0 1 1 * *
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0 0 * 1 0 0 * 1
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0 1 * * 0 1 * *
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"""
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def depth_first_search(grid: list[list[int]], row: int, col: int, visit: set) -> int:
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"""
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Recursive Backtracking Depth First Search Algorithm
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Starting from top left of a matrix, count the number of
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paths that can reach the bottom right of a matrix.
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1 represents a block (inaccessible)
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0 represents a valid space (accessible)
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0 0 0 0
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1 1 0 0
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0 0 0 1
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0 1 0 0
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>>> grid = [[0, 0, 0, 0], [1, 1, 0, 0], [0, 0, 0, 1], [0, 1, 0, 0]]
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>>> depth_first_search(grid, 0, 0, set())
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2
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0 0 0 0 0
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0 1 1 1 0
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0 1 1 1 0
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0 0 0 0 0
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>>> grid = [[0, 0, 0, 0, 0], [0, 1, 1, 1, 0], [0, 1, 1, 1, 0], [0, 0, 0, 0, 0]]
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>>> depth_first_search(grid, 0, 0, set())
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2
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"""
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row_length, col_length = len(grid), len(grid[0])
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if (
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min(row, col) < 0
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or row == row_length
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or col == col_length
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or (row, col) in visit
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or grid[row][col] == 1
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):
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return 0
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if row == row_length - 1 and col == col_length - 1:
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return 1
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visit.add((row, col))
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count = 0
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count += depth_first_search(grid, row + 1, col, visit)
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count += depth_first_search(grid, row - 1, col, visit)
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count += depth_first_search(grid, row, col + 1, visit)
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count += depth_first_search(grid, row, col - 1, visit)
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visit.remove((row, col))
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return count
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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