Available Captures for Rook LeetCode solution in Python

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class Solution:
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    def numRookCaptures(self, board: List[List[str]]) -> int:
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        # Checking for possible case to the right of Rook
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        def right_position(n, i):
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            List = i[0:n][::-1]  # taking list to the right of rook
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            pIndex, bIndex = -1, -1
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            if "p" in List:  # Checking if  'p' in list
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                pIndex = List.index("p")
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            if "B" in List:  # Checking if 'B' in list
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                bIndex = List.index("B")
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            print(bIndex, pIndex, List)
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            if bIndex == -1 and pIndex > -1:  # if list does not have 'B' and have 'p'
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                return True
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            if pIndex == -1:  # if list does not have 'p'
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                return False
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            return bIndex > pIndex
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        def left_position(n, i):
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            List = i[n + 1 :]  # taking list to the right of rook
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            pIndex, bIndex = -1, -1
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            if "p" in List:
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                pIndex = List.index("p")
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            if "B" in List:
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                bIndex = List.index("B")
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            print(bIndex, pIndex, List)
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            if bIndex == -1 and pIndex > -1:
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                return True
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            if pIndex == -1:
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                return False
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            return bIndex > pIndex
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        Count = 0
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        # Checking for possibilites in row
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        for i in board:
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            if "R" in i:
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                print(i)
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                n = i.index("R")
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                if left_position(n, i):
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                    Count += 1
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                if right_position(n, i):
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                    Count += 1
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        Col = []
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        # checking for possibilites in col
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        for i in range(0, len(board)):
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            Col.append(board[i][n])  # taking the elements from the same col of Rook
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        n = Col.index("R")
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        if left_position(n, Col):
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            Count += 1
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        if right_position(n, Col):
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            Count += 1
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        return Count