# 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.

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` |

## Python 3

`# Python 3 program to find k'th ` `# non-repeating character in a string` `MAX_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 code` `if` `__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#

`// C# program to find k'th non-repeating` `// character in a string` `using` `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.` |

## Javascript

`<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 program` `let 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>` |

**Output:**

k'th non-repeating character is r

**Time Complexity:** O(m log m) , here m is the MAX_CHARS = 256**Auxiliary Space:** O(m), here m is the MAX_CHARS = 256

**Space Optimized Solution : **

This can be space optimized and can be solved using single index array only. Below is the space optimized solution:

## C++

`#include <bits/stdc++.h>` `using` `namespace` `std;` `#define MAX_CHAR 256` `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[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` |

## Java

`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` |

## Python3

`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 ` `=` `3` `res ` `=` `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` |

## C#

`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` |

## Javascript

`<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>` |

**Output:**

k'th non-repeating character is r

**Time Complexity:** O(m log m) , here m is the MAX_CHARS = 256**Auxiliary Space:** O(m), here m is the MAX_CHARS = 256

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

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