K'th Non-repeating Character - GeeksforGeeks

18 Nov, 2021
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(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.
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] = iIf count[x] is 2, then remove x from index[], i.e., index[x] = nNow 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[].
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] = iIf count[x] is 2, then remove x from index[], i.e., index[x] = n
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++
Java
Python 3
C#
Javascript
// 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 codeint 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 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
# Python 3 program to find k'th# non-repeating character in a stringMAX_CHAR = 256 # Returns index of k'th non-repeating# character in given string str[]def kthNonRepeating(str, k):     n = len(str)     # count[x] is going to store count of     # character 'x' in str. If x is not    # present, then it is going to store 0.    count = [0] * 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[]    index = [0] * MAX_CHAR     # Initialize counts of all characters    # and indexes of non-repeating characters.    for i in range( MAX_CHAR):        count[i] = 0        index[i] = n # A value more than any                     # index in str[]     # Traverse the input string    for i in range(n):                 # Find current character and        # increment its count        x = str[i]        count[ord(x)] += 1         # If this is first occurrence, then        # set value in index as index of it.        if (count[ord(x)] == 1):            index[ord(x)] = i         # If character repeats, then remove        # it from index[]        if (count[ord(x)] == 2):            index[ord(x)] = n     # Sort index[] in increasing order. This step    # takes O(1) time as size of index is 256 only    index.sort()     # After sorting, if index[k-1] is value,    # then return it, else return -1.    return index[k - 1] if (index[k - 1] != n) else -1 # Driver codeif __name__ == "__main__":    str = "geeksforgeeks"    k = 3    res = kthNonRepeating(str, k)    if(res == -1):        print("There are less than k",              "non-repeating characters")    else:        print("k'th non-repeating character is",                                       str[res]) # This code is contributed# by ChitraNayal
// C# program to find k'th non-repeating// character in a stringusing System; 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;                         // A value more than any index            // in str[]            index[i] = n;        }         // 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        Array.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 str = "geeksforgeeks";        int k = 3;        int res = kthNonRepeating(str, k);             Console.Write(res == -1 ? "There are less"         + " than k non-repeating characters" :            "k'th non-repeating character is "                                   + str[res]);    }} // This code is contributed by nitin mittal.
<script>// Javascript program to find k'th non-repeating character// in a string let MAX_CHAR = 256; // Returns index of k'th non-repeating character in    // given string str[]function kthNonRepeating(str,k){    let 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.        let count = new Array(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[]        let index = new Array(MAX_CHAR);           // Initialize counts of all characters and indexes        // of non-repeating  characters.        for (let 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 (let i = 0; i < n; i++)        {            // Find current character and increment its            // count            let x = str[i];            ++count[x.charCodeAt(0)];               // If this is first occurrence, then set value            // in index as index of it.            if (count[x.charCodeAt(0)] == 1)                index[x.charCodeAt(0)] = i;               // If character repeats, then remove it from            // index[]            if (count[x.charCodeAt(0)] == 2)                index[x.charCodeAt(0)] = n;        }           // Sort index[] in increasing order.  This step        // takes O(1) time as size of index is 256 only        (index).sort(function(a,b){return a-b;});           // After sorting, if index[k-1] is value, then        // return it, else return -1.        return (index[k-1] != n)? index[k-1] : -1;} // driver programlet str = "geeksforgeeks";let k = 3;let res = kthNonRepeating(str, k);document.write(res == -1 ? "There are less than k non-repeating characters" :                           "k'th non-repeating character is  " + str[res]); // This code is contributed by ab2127</script>
k'th non-repeating character is  r
 
Space Optimized Solution : This can be space optimized and can be solved using single index array only. Below is the space optimized solution: 
C++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std;#define MAX_CHAR 256int kthNonRepeating(string input, int k){    int inputLength = input.length();     /*     * indexArr will store index of non-repeating     * characters, inputLength for characters not in input     * and inputLength+1 for repeated characters.     */    int indexArr[MAX_CHAR];     // initialize all values in indexArr as inputLength.    for (int i = 0; i < MAX_CHAR; i++) {        indexArr[i] = inputLength;    }    for (int i = 0; i < inputLength; i++) {        char c = input[i];        if (indexArr == inputLength) {            indexArr = i;        }        else {            indexArr = inputLength + 2;        }    }     sort(indexArr, indexArr + MAX_CHAR);    return (indexArr[k - 1] != inputLength)               ? indexArr[k - 1]               : -1;} int main(){    string input = "geeksforgeeks";    int k = 3;    int res = kthNonRepeating(input, k);    if (res == -1)        cout << "There are less than k non-repeating "                "characters";    else        cout << "k'th non-repeating character is  "             << input[res];    return 0;} // This code is contributed by gauravrajput1
import java.util.*; public class GFG {    public static int MAX_CHAR = 256;         public static void main (String[] args)     {        final String input = "geeksforgeeks";        int k = 3;        int res = kthNonRepeating(input, k);                   System.out.println(res == -1 ? "There are less than k non-repeating characters" :                           "k'th non-repeating character is  " + input.charAt(res));    }     public static int kthNonRepeating(final String input, final int k) {        final int inputLength = input.length();         /*         * indexArr will store index of non-repeating characters,         * inputLength for characters not in input and         * inputLength+1 for repeated characters.         */        final int[] indexArr = new int[MAX_CHAR];                 // initialize all values in indexArr as inputLength.        Arrays.fill(indexArr, inputLength);                 for (int i = 0; i < inputLength ; i++) {            final char c = input.charAt(i);            if (indexArr == inputLength) {                indexArr = i;            } else {                indexArr = inputLength + 2;            }        }                 Arrays.sort(indexArr);                 return (indexArr[k-1] != inputLength) ? indexArr[k-1] : -1;    }}// Contributed by AK
MAX_CHAR = 256 def kthNonRepeating(Input,k):    inputLength = len(Input)         # indexArr will store index of non-repeating characters,    # inputLength for characters not in input and    # inputLength+1 for repeated characters.         # initialize all values in indexArr as inputLength.    indexArr = [inputLength for i in range(MAX_CHAR)]         for i in range(inputLength):        c = Input[i]        if (indexArr[ord(c)] == inputLength):            indexArr[ord(c)] = i        else:            indexArr[ord(c)] = inputLength + 2    indexArr.sort()    if(indexArr[k - 1] != inputLength):        return indexArr[k - 1]    else:        return -1 Input = "geeksforgeeks"k = 3res = kthNonRepeating(Input, k)if(res == -1):    print("There are less than k non-repeating characters")else:    print("k'th non-repeating character is", Input[res])     # This code is contributed by rag2127
using System; public class GFG{    public static int MAX_CHAR = 256;    static public void Main ()    {        string input = "geeksforgeeks";         int k = 3;         int res = kthNonRepeating(input, k);                      Console.WriteLine(res == -1 ? "There are less than k non-repeating characters" :                            "k'th non-repeating character is  " + input[res]);    }         public static int kthNonRepeating(string input, int k) {        int inputLength = input.Length;           /*         * indexArr will store index of non-repeating characters,         * inputLength for characters not in input and         * inputLength+1 for repeated characters.         */        int[] indexArr = new int[MAX_CHAR];                   // initialize all values in indexArr as inputLength.        Array.Fill(indexArr, inputLength);                   for (int i = 0; i < inputLength ; i++) {            char c = input[i];            if (indexArr == inputLength) {                indexArr = i;            } else {                indexArr = inputLength + 2;            }        }                   Array.Sort(indexArr);        return (indexArr[k - 1] != inputLength) ? indexArr[k - 1] : -1;     }} // This code is contributed by avanitrachhadiya2155
<script> let MAX_CHAR = 256; function kthNonRepeating(input, k){    let inputLength = input.length;          /*         * indexArr will store index of non-repeating characters,         * inputLength for characters not in input and         * inputLength+1 for repeated characters.         */        let indexArr = new Array(MAX_CHAR);                  // initialize all values in indexArr as inputLength.        for(let i=0;i<MAX_CHAR;i++)        {            indexArr[i]=inputLength;        }                  for (let i = 0; i < inputLength ; i++) {            let c = input[i];            if (indexArr == inputLength) {                indexArr = i;            } else {                indexArr = inputLength + 2;            }        }                  (indexArr).sort(function(a,b){return a-b;});                  return (indexArr[k-1] != inputLength) ? indexArr[k-1] : -1;} let input = "geeksforgeeks";let k = 3;let res = kthNonRepeating(input, k);document.write(res == -1 ? "There are less than k non-repeating characters" :                           "k'th non-repeating character is  " + input[res]); // This code is contributed by unknown2108</script>
k'th non-repeating character is  r
 
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nitin mittal
ukasp
abhishekkumarnitrr
avanitrachhadiya2155
rag2127
ab2127
unknown2108
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