K’th Non-repeating Character

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.
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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)

1. Create an empty hash.
2. Scan input string from left to right and insert values and their counts in the hash.
3. 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.
```
1. Initialize all values in count[] as 0 and all values in index[] as n where n is length of string.
2. 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
3. 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.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.7 Average Difficulty : 2.7/5.0
Based on 44 vote(s)