CODEWITHSUNDEEP
pythonstringchar
StackOoverflow
StackOoverflow

Reputation: 35

How to write a code to determine if a character is in a string or not?

Here in this quiz it tells me to write a code that make sure if a charachter is in a string which is sorted alphabetically and it tells me to bisection search to determine if a character is in a string and tells me to write a recursive code and not to use "in" (for i in ...)

My code was like that :

def isIn(char, aStr):
    '''
    char: a single character
    aStr: an alphabetized string

    returns: True if char is in aStr; False otherwise
    '''
    var = 0
    if char == '' and aStr == '':
        return True
    elif char == '' or aStr == '' :
        return False
    elif aStr[int((len(aStr))/2)] == char :
        return True
    elif aStr[int((len(aStr))/2)] < char :
        var = int((len(aStr))/2)
        return isIn(char , aStr[var:])
    elif aStr[int((len(aStr))/2)] > char :
        var = int((len(aStr))/2)
        return isIn(char , aStr[0 : var])

but when i pass values like (char = 'o' aStr = 'efggnqu', or char = 'y' aStr = 'orstv',or char = 'y' aStr = 'aamotv') it gives me an error like this :

Traceback (most recent call last):
  File "submission.py", line 21, in isIn
    return isIn(char , aStr[0 : var])
  File "submission.py", line 18, in isIn
    return isIn(char , aStr[var:])
  File "submission.py"...OUTPUT TRUNCATED

what is the solution?

Upvotes: 2

Views: 253

Answers (1)

anonymoose
anonymoose

Reputation: 868

Here's one fix:

def isIn(char, aStr):
    var = len(aStr) // 2 #Pulled out for readability (and speed)
    if char == '' and aStr == '':
        return True
    elif char == '' or aStr == '' :
        return False
    elif aStr[var] == char:
        return True
    elif aStr[var] < char:
        return isIn(char, aStr[var + 1:])
    elif aStr[var] > char:
        return isIn(char, aStr[:var])

The crucial piece is the +1 on the third line up from the bottom. Without it, a single character is passed over and over again. 1 // 2 rounds down to 0, and "n"[0:] is "n" again.

Sample output:

>>> isIn("o", "efgnqu")
False
>>> isIn("y", "orstv")
False
>>> isIn("y", "aamotv")
False
>>>

Upvotes: 3

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