can you please convert this python code into java? Python code is as shown below: #recursive function def row_puzzle_rec(row, pos, visited): # if the element at the current position is o we have reached our goal if row[pos] == 0: possible = True else: # make a copy of the visited array visited visited[:] # if the element at the current position has been already visited then it's a loop. # as then its not possible to reach the last element, set possible to False if visited[pos]: possible = False else: #set visitied for the element as True visited[pos] = True possible = False # if its possible to move left then recurse in the left direction if pos - row[pos]>o and row_puzzle_rec(row, pos - row[pos], visited): # return immediately if the goal is reached return True # if its possible to move right then recurse in the right direction if pos + row[pos] < len(row) and row_puzzle_rec(row, pos + row[pos], visited): # return immediately if the goal is reached. return True return possible def row_puzzle(row): #visited array will be used to check if a particular element was visited # initialize the array to False visited - [False for item in range(len(row))] #second argument is the token position initialized to o return row_puzzle_rec(row, o, visited) possible row_puzzle([2, 4, 5, 3, 1, 3, 1, 4, 0)) print(possible) possible = row_puzzle([1, 3, 2, 1, 3, 4, ol) print(possible)
can you please convert this python code into java? Python code is as shown below: #recursive function def row_puzzle_rec(row, pos, visited): # if the element at the current position is o we have reached our goal if row[pos] == 0: possible = True else: # make a copy of the visited array visited visited[:] # if the element at the current position has been already visited then it's a loop. # as then its not possible to reach the last element, set possible to False if visited[pos]: possible = False else: #set visitied for the element as True visited[pos] = True possible = False # if its possible to move left then recurse in the left direction if pos - row[pos]>o and row_puzzle_rec(row, pos - row[pos], visited): # return immediately if the goal is reached return True # if its possible to move right then recurse in the right direction if pos + row[pos] < len(row) and row_puzzle_rec(row, pos + row[pos], visited): # return immediately if the goal is reached. return True return possible def row_puzzle(row): #visited array will be used to check if a particular element was visited # initialize the array to False visited - [False for item in range(len(row))] #second argument is the token position initialized to o return row_puzzle_rec(row, o, visited) possible row_puzzle([2, 4, 5, 3, 1, 3, 1, 4, 0)) print(possible) possible = row_puzzle([1, 3, 2, 1, 3, 4, ol) print(possible)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Answer 12....
![can you please convert this python code into java?
Python code is as shown below:
# recursive function
def row_puzzle_rec(row, pos, visited):
# if the element at the current position is o we have reached
our goal
if row[pos] == 0:
possible = True
else:
# make a copy of the visited array
visited visited[:]
# if the element at the current position has been already
visited then it's a loop.
# as then its not possible to reach the last element, set
possible to False
if visited[pos]:
possible = False
else:
#set visitied for the element as True
visited[pos] = True
possible = False
# if its possible to move left then recurse in the left
direction
if pos - row[pos]>o and row_puzzle_rec(row, pos -
row[pos], visited):
# return immediately if the goal is reached
return True
# if its possible to move right then recurse in the right
direction
if pos + row[pos] < len(row) and row_puzzle_rec(row, pos +
row[pos], visited):
# return immediately if the goal is reached.
return True
return possible
def row_puzzle(row):
#visited array will be used to check if a particular element was
visited
# initialize the array to False
visited = [False for item in range(len(row))]
#second argument is the token position initialized to o
return row_puzzle_rec(row, o, visited)
possible = row_puzzle([2, 4, 5, 3, 1, 3, 1, 4, ol)
print(possible)
possible = row_puzzle([1, 3, 2, 1, 3, 4, 0])
print(possible)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff8bc427-4dd5-458f-a0de-0d1350ec468a%2Fe7d98e7d-6879-4d1b-b807-1f98255937db%2Fwz8k9p7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:can you please convert this python code into java?
Python code is as shown below:
# recursive function
def row_puzzle_rec(row, pos, visited):
# if the element at the current position is o we have reached
our goal
if row[pos] == 0:
possible = True
else:
# make a copy of the visited array
visited visited[:]
# if the element at the current position has been already
visited then it's a loop.
# as then its not possible to reach the last element, set
possible to False
if visited[pos]:
possible = False
else:
#set visitied for the element as True
visited[pos] = True
possible = False
# if its possible to move left then recurse in the left
direction
if pos - row[pos]>o and row_puzzle_rec(row, pos -
row[pos], visited):
# return immediately if the goal is reached
return True
# if its possible to move right then recurse in the right
direction
if pos + row[pos] < len(row) and row_puzzle_rec(row, pos +
row[pos], visited):
# return immediately if the goal is reached.
return True
return possible
def row_puzzle(row):
#visited array will be used to check if a particular element was
visited
# initialize the array to False
visited = [False for item in range(len(row))]
#second argument is the token position initialized to o
return row_puzzle_rec(row, o, visited)
possible = row_puzzle([2, 4, 5, 3, 1, 3, 1, 4, ol)
print(possible)
possible = row_puzzle([1, 3, 2, 1, 3, 4, 0])
print(possible)
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