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 = neveropen, k = 3
Output : r
First non-repeating character is f,
second is o and third is r.
Input : str = neveropen, k = 2
Output : o
Input : str = neveropen, 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.
Implementation:
Python3
# Function to find k'th non repeating character# in stringfrom collections import OrderedDictdef 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 functionif __name__ == "__main__": input = "neveropen" k = 3 print (kthRepeating(input, k)) |
Output
r
This article is contributed by Shashank Mishra (Gullu). If you like Lazyroar and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the Lazyroar main page and help other Geeks.
