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Program to find all possible strings typed using phone keypad in python
Suppose we have a string containing digits from 2-9. We have to find all possible letter combinations that the number could generate. One mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does map some characters but no letters.
| 1 |
2 a b c |
3 d e f |
| 4 g h i |
5 j k l |
6 m n o |
| 7 p q r s |
8 t u v |
9 w x y z |
| * |
0 |
# |
For an example, if the given string is “49”, then the possible strings will be ['gw', 'gx', 'gy', 'gz', 'hw', 'hx', 'hy', 'hz', 'iw', 'ix', 'iy', 'iz']
To solve this, we will follow these steps:
- Define an array called solve to solve the problem recursively
- solve method takes digits, characters, result, current_string and current_level, the function will be like
- if current_level = length of digits, then add current string after the result, and return
- for all characters i in characters[digits[current_level]]
- perform solve(digits, characters, result, current_string + i, current_level + 1)
- The actual function will be like
- if digits length is 0, then return empty list
- define one map to hold numbers and corresponding characters as a string
- result := an empty list
- call solve(digits, characters, result, “”, 0)
Let us see the following implementation to get better understanding:
Example
class Solution(object):
def letterCombinations(self, digits):
if len(digits) == 0:
return []
characters = {2:"abc",3:"def",4:"ghi",5:"jkl",6:"mno",7:"pqrs",8:"tuv",9:"wxyz"}
result = []
self.solve(digits,characters,result)
return result
def solve(self, digits, characters, result, current_string="",current_level = 0):
if current_level == len(digits):
result.append(current_string)
return
for i in characters[int(digits[current_level])]:
self.solve(digits,characters,result,current_string+i,current_level+1)
ob1 = Solution()
print(ob1.letterCombinations("49"))
Input
"49"
Output
['gw', 'gx', 'gy', 'gz', 'hw', 'hx', 'hy', 'hz', 'iw', 'ix', 'iy', 'iz']
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