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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