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 is mainly an extension of First non-repeating character problem.

**Method 1 (Simple : O(n ^{2})**

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.

## C++

// C++ program to find k'th non-repeating character // in a string #include <bits/stdc++.h> using namespace std; const int MAX_CHAR = 256; // Returns index of k'th non-repeating character in // given string str[] int kthNonRepeating(string str, int k) { int n = str.length(); // count[x] is going to store count of // character 'x' in str. If x is not present, // then it is going to store 0. int count[MAX_CHAR]; // index[x] is going to store index of character // 'x' in str. If x is not present or x is // repeating, then it is going to store a value // (for example, length of string) that cannot be // a valid index in str[] int index[MAX_CHAR]; // Initialize counts of all characters and indexes // of non-repeating characters. for (int i = 0; i < MAX_CHAR; i++) { count[i] = 0; index[i] = n; // A value more than any index // in str[] } // Traverse the input string for (int i = 0; i < n; i++) { // Find current character and increment its // count char x = str[i]; ++count[x]; // If this is first occurrence, then set value // in index as index of it. if (count[x] == 1) index[x] = i; // If character repeats, then remove it from // index[] if (count[x] == 2) index[x] = n; } // Sort index[] in increasing order. This step // takes O(1) time as size of index is 256 only sort(index, index+MAX_CHAR); // After sorting, if index[k-1] is value, then // return it, else return -1. return (index[k-1] != n)? index[k-1] : -1; } // Driver code int main() { string str = "geeksforgeeks"; int k = 3; int res = kthNonRepeating(str, k); (res == -1)? cout << "There are less than k non-" "repeating characters" : cout << "k'th non-repeating character" " is "<< str[res]; return 0; }

## Java

// Java program to find k'th non-repeating character // in a string import java.util.Arrays; class GFG { public static int MAX_CHAR = 256; // Returns index of k'th non-repeating character in // given string str[] static int kthNonRepeating(String str, int k) { int n = str.length(); // count[x] is going to store count of // character 'x' in str. If x is not present, // then it is going to store 0. int[] count = new int[MAX_CHAR]; // index[x] is going to store index of character // 'x' in str. If x is not present or x is // repeating, then it is going to store a value // (for example, length of string) that cannot be // a valid index in str[] int[] index = new int[MAX_CHAR]; // Initialize counts of all characters and indexes // of non-repeating characters. for (int i = 0; i < MAX_CHAR; i++) { count[i] = 0; index[i] = n; // A value more than any index // in str[] } // Traverse the input string for (int i = 0; i < n; i++) { // Find current character and increment its // count char x = str.charAt(i); ++count[x]; // If this is first occurrence, then set value // in index as index of it. if (count[x] == 1) index[x] = i; // If character repeats, then remove it from // index[] if (count[x] == 2) index[x] = n; } // Sort index[] in increasing order. This step // takes O(1) time as size of index is 256 only Arrays.sort(index); // After sorting, if index[k-1] is value, then // return it, else return -1. return (index[k-1] != n)? index[k-1] : -1; } // driver program public static void main (String[] args) { String str = "geeksforgeeks"; int k = 3; int res = kthNonRepeating(str, k); System.out.println(res == -1 ? "There are less than k non-repeating characters" : "k'th non-repeating character is " + str.charAt(res)); } } // Contributed by Pramod Kumar

Output :

k'th non-repeating character is r

This article is contributed by Shivam Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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