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 .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) |
Output:
r
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