K’th Non-repeating Character in Python using List Comprehension and OrderedDict

Given a string and a number k, find the k-th non-repeating character in the string. Consider a large input string with lacs of characters and a small character set. How to find the character by only doing only one traversal of input string?

Examples:

Input : str = geeksforgeeks, k = 3
Output : r
First non-repeating character is f,
second is o and third is r.

Input : str = geeksforgeeks, k = 2
Output : o

Input : str = geeksforgeeks, k = 4
Output : Less than k non-repeating
         characters in input.



This problem has existing solution please refer link. We can solve this problem quickly in python using List Comprehension and OrderedDict.

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# Function to find k'th non repeating character 
# in string
from collections import OrderedDict
  
def kthRepeating(input,k):
  
    # OrderedDict returns a dictionary data 
        # structure having characters of input
    # string as keys in the same order they 
        # were inserted and 0 as their default value
    dict=OrderedDict.fromkeys(input,0)
  
    # now traverse input string to calculate
        # frequency of each character
    for ch in input:
        dict[ch]+=1
  
    # now extract list of all keys whose value
        # is 1 from dict Ordered Dictionary 
    nonRepeatDict = [key for (key,value) in \
                    dict.iteritems() if value==1]
      
    # now return (k-1)th character from above list
    if len(nonRepeatDict) < k:
        return 'Less than k non-repeating \
                characters in input.'
    else:
        return nonRepeatDict[k-1]
  
# Driver function
if __name__ == "__main__":
    input = "geeksforgeeks"
    k = 3
    print kthRepeating(input, k)

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Output:


r



This article is contributed by Shashank Mishra (Gullu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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