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K’th Non-repeating Character in Python using List Comprehension and OrderedDict
  • Difficulty Level : Medium
  • Last Updated : 23 Nov, 2020

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.




# 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.items() 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)) 

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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