I was wondering which kind of O notation this code is, can somebody help me to figure out? I wrote it but when I was interrogated about which kind of O notation is, and the only thing I could say it was linear but I have the feeling that recursion+iteration should be exponential?
listPlindromes =  def palindrome( givenString, n ): if n == 1: #print givenString return None else: #print givenString[:n] #print ('forward: '+givenString[:n] +' backwards: '+givenString[:n][::-1]) #Get difference between lengths lenDifference = len(givenString) - len(givenString[:n]) #If there is a difference means there at least one more word/palindrome could exist #therefore it need to be tested if lenDifference > 0: for xTest in range(0,lenDifference-1) : newWord=givenString[xTest:n+xTest] if newWord == newWord[::-1]: if len(newWord) > 0: listPlindromes.append(newWord) #print newWord else: if givenString[:n] == givenString[:n][::-1]: listPlindromes.append(givenString[:n]) #print('palindrome: '+givenString) return palindrome(givenString,n-1) givenString='osooso' palindrome(givenString, len(givenString)) print(listPlindromes)
The code you've got there runs in linear time; that is, its primary bottleneck is how large the string you give it is.
That said, there are some things that can be improved:
- You only need to pass in the string
- A string of length 1 is a palindrome
- There's a lot of unnecessary/unneeded checks for length; simply slicing the string in half would be sufficient to start your checks
As for the iterative + recursive approach, it really depends on how it's authored, but I have a recursive approach in mind which could be O(n) as well.
- Start with the whole string.
- If the two ends match:
- If the length of one side* of the string is greater than 1, take a slice of it from that position up to 1 less than the last character.
- Invoke function.
- If the two ends match and the length of one side* of the string is less than or equal to 1, we're done - return true.
- Return false, as we've found a mismatch.
*: Regardless if the string is even or odd, cutting it in half will give you an even number of characters on both sides.