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

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  • Difficulty Level : Easy
  • 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?


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 
    # now traverse input string to calculate 
        # frequency of each character 
    for ch in input
    # 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.' 
        return nonRepeatDict[k-1
# Driver function 
if __name__ == "__main__"
    input = "geeksforgeeks"
    k = 3
    print (kthRepeating(input, k)) 



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