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3

$begingroup$

Task

Old mobile phones had the ability to type characters by pressing a number. The letter a could be typed by pressing 2 once. The letter b could be typed by pressing 2 twice.

Given a sequence of numbers, give all possible letter combinations.

For example: The number 23 could give an output ad, ae, af, bd, be, bf, cd, ce, cf

My recursive solution to this problem is given below.

def num_to_char(value):

    if value == 2: return ["a","b","c"]

    if value == 3: return ["d","e","f"]

    if value == 4: return ["g","h","i"]

    if value == 5: return ["j","k","l"]

    if value == 6: return ["m","n","o"]

    if value == 7: return ["p","q","r","s"]

    if value == 8: return ["t","u","v"]

    if value == 9: return ["w","x","y","z"]



def convert_num(number, current_string = ""):

    if number == []:

        print(current_string)

        return 

    get_list = num_to_char(int(number[0]))

    for character in get_list:

        current_string += character

        convert_num(number[1:], current_string)

        current_string = current_string[:-1]



num_to_covert = list("234")

convert_num(num_to_covert)

python python-3.x programming-challenge

share|improve this question

edited 8 hours ago

200_success

133k20162432

asked 8 hours ago

EMLEML

3167

$endgroup$

add a comment | 

3

$begingroup$

Task

Old mobile phones had the ability to type characters by pressing a number. The letter a could be typed by pressing 2 once. The letter b could be typed by pressing 2 twice.

Given a sequence of numbers, give all possible letter combinations.

For example: The number 23 could give an output ad, ae, af, bd, be, bf, cd, ce, cf

My recursive solution to this problem is given below.

def num_to_char(value):

    if value == 2: return ["a","b","c"]

    if value == 3: return ["d","e","f"]

    if value == 4: return ["g","h","i"]

    if value == 5: return ["j","k","l"]

    if value == 6: return ["m","n","o"]

    if value == 7: return ["p","q","r","s"]

    if value == 8: return ["t","u","v"]

    if value == 9: return ["w","x","y","z"]



def convert_num(number, current_string = ""):

    if number == []:

        print(current_string)

        return 

    get_list = num_to_char(int(number[0]))

    for character in get_list:

        current_string += character

        convert_num(number[1:], current_string)

        current_string = current_string[:-1]



num_to_covert = list("234")

convert_num(num_to_covert)

python python-3.x programming-challenge

share|improve this question

edited 8 hours ago

200_success

133k20162432

asked 8 hours ago

EMLEML

3167

$endgroup$

add a comment | 

$begingroup$

Task

Old mobile phones had the ability to type characters by pressing a number. The letter a could be typed by pressing 2 once. The letter b could be typed by pressing 2 twice.

Given a sequence of numbers, give all possible letter combinations.

For example: The number 23 could give an output ad, ae, af, bd, be, bf, cd, ce, cf

My recursive solution to this problem is given below.

def num_to_char(value):

    if value == 2: return ["a","b","c"]

    if value == 3: return ["d","e","f"]

    if value == 4: return ["g","h","i"]

    if value == 5: return ["j","k","l"]

    if value == 6: return ["m","n","o"]

    if value == 7: return ["p","q","r","s"]

    if value == 8: return ["t","u","v"]

    if value == 9: return ["w","x","y","z"]



def convert_num(number, current_string = ""):

    if number == []:

        print(current_string)

        return 

    get_list = num_to_char(int(number[0]))

    for character in get_list:

        current_string += character

        convert_num(number[1:], current_string)

        current_string = current_string[:-1]



num_to_covert = list("234")

convert_num(num_to_covert)

python python-3.x programming-challenge

share|improve this question

edited 8 hours ago

200_success

133k20162432

asked 8 hours ago

EMLEML

3167

$endgroup$

def num_to_char(value):

    if value == 2: return ["a","b","c"]

    if value == 3: return ["d","e","f"]

    if value == 4: return ["g","h","i"]

    if value == 5: return ["j","k","l"]

    if value == 6: return ["m","n","o"]

    if value == 7: return ["p","q","r","s"]

    if value == 8: return ["t","u","v"]

    if value == 9: return ["w","x","y","z"]



def convert_num(number, current_string = ""):

    if number == []:

        print(current_string)

        return 

    get_list = num_to_char(int(number[0]))

    for character in get_list:

        current_string += character

        convert_num(number[1:], current_string)

        current_string = current_string[:-1]



num_to_covert = list("234")

convert_num(num_to_covert)

share|improve this question

edited 8 hours ago

200_success

133k20162432

asked 8 hours ago

EMLEML

3167

share|improve this question

edited 8 hours ago

200_success

133k20162432

asked 8 hours ago

EMLEML

3167

edited 8 hours ago

200_success

133k20162432

edited 8 hours ago

200_success

133k20162432

asked 8 hours ago

EMLEML

3167

asked 8 hours ago

EMLEML

3167

1 Answer
1

active

oldest

votes

4

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product



KEYPAD = {

                 '2': 'abc',  '3': 'def',

    '4': 'ghi',  '5': 'jkl',  '6': 'mno',

    '7': 'pqrs', '8': 'tuv',  '9': 'wxyz',

}



def convert_num(number):

    letters = [KEYPAD[c] for c in number]

    return [''.join(combo) for combo in product(*letters)]



print(convert_num('234'))

share|improve this answer

answered 7 hours ago

200_success200_success

133k20162432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant. Sadly I actually had no idea what a cartesian product was so thanks for educating me :)
  $endgroup$
  – EML
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  In general, any time you want to do some kind of fancy iteration in Python, look at itertools first.
  $endgroup$
  – 200_success
  7 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product



KEYPAD = {

                 '2': 'abc',  '3': 'def',

    '4': 'ghi',  '5': 'jkl',  '6': 'mno',

    '7': 'pqrs', '8': 'tuv',  '9': 'wxyz',

}



def convert_num(number):

    letters = [KEYPAD[c] for c in number]

    return [''.join(combo) for combo in product(*letters)]



print(convert_num('234'))

share|improve this answer

answered 7 hours ago

200_success200_success

133k20162432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant. Sadly I actually had no idea what a cartesian product was so thanks for educating me :)
  $endgroup$
  – EML
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  In general, any time you want to do some kind of fancy iteration in Python, look at itertools first.
  $endgroup$
  – 200_success
  7 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product



KEYPAD = {

                 '2': 'abc',  '3': 'def',

    '4': 'ghi',  '5': 'jkl',  '6': 'mno',

    '7': 'pqrs', '8': 'tuv',  '9': 'wxyz',

}



def convert_num(number):

    letters = [KEYPAD[c] for c in number]

    return [''.join(combo) for combo in product(*letters)]



print(convert_num('234'))

share|improve this answer

answered 7 hours ago

200_success200_success

133k20162432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant. Sadly I actually had no idea what a cartesian product was so thanks for educating me :)
  $endgroup$
  – EML
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  In general, any time you want to do some kind of fancy iteration in Python, look at itertools first.
  $endgroup$
  – 200_success
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product



KEYPAD = {

                 '2': 'abc',  '3': 'def',

    '4': 'ghi',  '5': 'jkl',  '6': 'mno',

    '7': 'pqrs', '8': 'tuv',  '9': 'wxyz',

}



def convert_num(number):

    letters = [KEYPAD[c] for c in number]

    return [''.join(combo) for combo in product(*letters)]



print(convert_num('234'))

share|improve this answer

answered 7 hours ago

200_success200_success

133k20162432

$endgroup$

from itertools import product



KEYPAD = {

                 '2': 'abc',  '3': 'def',

    '4': 'ghi',  '5': 'jkl',  '6': 'mno',

    '7': 'pqrs', '8': 'tuv',  '9': 'wxyz',

}



def convert_num(number):

    letters = [KEYPAD[c] for c in number]

    return [''.join(combo) for combo in product(*letters)]



print(convert_num('234'))

share|improve this answer

answered 7 hours ago

200_success200_success

133k20162432

answered 7 hours ago

200_success200_success

133k20162432

answered 7 hours ago

200_success200_success

133k20162432