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 is mainly an extension of First non-repeating character problem.
Method 1 (Simple : O(n2)
A Simple Solution is to run two loops. Start traversing from left side. For every character, check if it repeats or not. If the character doesn’t repeat, increment count of non-repeating characters. When the count becomes k, return the character.
Method 2 (O(n) but requires two traversals)
- Create an empty hash.
- Scan input string from left to right and insert values and their counts in the hash.
- Scan input string from left to right and keep count of characters with counts more than 1. When count becomes k, return the character.
Method 3 (O(n) and requires one traversal)
The idea is to use two auxiliary arrays of size 256 (Assuming that characters are stored using 8 bits). The two arrays are:
count[x] : Stores count of character 'x' in str. If x is not present, then it stores 0. index[x] : Stores indexes of non-repeating characters in str. If a character 'x' is not present or x is repeating, then it stores a value that cannot be a valid index in str. For example, length of string.
- Initialize all values in count as 0 and all values in index as n where n is length of string.
- Traverse the input string str and do following for every character c = str[i].
- Increment count[x].
- If count[x] is 1, then store index of x in index[x], i.e., index[x] = i
- If count[x] is 2, then remove x from index, i.e., index[x] = n
- Now index has indexes of all non-repeating characters. Sort index in increasing order so that we get k’th smallest element at index[k]. Note that this step takes O(1) time because there are only 256 elements in index.
Below is implementation of above idea.
k'th non-repeating character is r
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