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.
# 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
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