# 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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. ` |

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**Output:**

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:

`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 ` |

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**Output:**

k'th non-repeating character is r

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