在两个单词之间找到一个共同的字母，找不到我的代码中的错误

Question

I'm supposed to write a code which gets two strings as an input and outputs True if there is a common letter and false if there isn't, I'm supposed to do this using recursion.

Running examples:

any_char_present("adsf","") #returns False
any_char_present("a","sfgserta") #returns true

My code gives me a correct answer for most cases, but for this case it gave me the wrong answer:

>>> any_char_present("abcefghijklmno","pqrstuvwxyza")
False

I can't figure out what's the bug in my code, here it is:

def any_char_present(s1,s2):
    if not s1 or not s2:
        return False
    if s1[0]==s2[0]:
        return True
    elif s1[0]!=s2[0]:
        s1=s1[1:]
        return any_char_present(s1,s2) or any_char_present(s2,s1)

Answer 1

This is because in your code, at the end of the first call, you are setting s1 to be: 'bcefghijklmno' and any comparisons made after that point will be different therefore it gives you False .

If this is a programming exercise, I will let you figure out the solution for yourself since that is probably why you are doing this exercise in the first place.

Other wise, you can do it as:

def any_char_present(s1, s2):
    return len(list(set(s1) & set(s2))) > 0

This will create a list of letters that are present in both strings and will then return True if the length of that list is greater than 0, ie there is atleast one common letter and False otherwise.

Hope that helps.

Answer 2

I will assume that's homework so I will only guide you in the right direction. Before that, just a thought: your code is not very efficient as you will test several times the same "subwords" to be convince, use a print s1,s2 before the first if:

>>> any_char_present("abc","def")
abc def
bc def
c def
 def
def
def c
ef c   
f c    #this
 c
c
c f    #is here
 f
f
c ef
 ef
ef
def bc
ef bc
f bc
 bc
bc
bc f
c f   #is here again
 f
f
f c   #and here 
 c
c
bc ef
c ef    
 ef
ef
ef c    
f c  #and also there
 c
c
c f #or there
 f
f

Now, for your problem: during the first iteration, you check s1[0] with s2[0 ], if they are different, you try to check the rest of s1 with s2 . But you haven't check s1[0] with the rest of s2 yet.

So if the first letter of s1 isn't the first letter of s2 , but is the second letter of s2 , your code will fail.

PS: if your homework isn't about recursion, check the in operator:

>>> 'c' in 'abcd'
True

Answer 3

I'd suggest a better and efficient solution as follows

def any_char_present(s1,s2):
  list1 = list(s1)  
  list2 = list(s2)  
  return set(list1).intersection(list2)

The last statement would return a list of chars which are common to both strings.