# Queries to find the first non-repeating character in the sub-string of a string

Given a string str, the task is to answer Q queries where every query consists of two integers L and R and we have to find the first non-repeating character in the sub-string str[L…R]. If there is no non-repeating character then print -1.

Examples:

Input: str = “GeeksForGeeks”, q[] = {{0, 3}, {2, 3}, {5, 12}}
Output:
G
e
F
Sub-string for the queries are “Geek”, “ek” and “ForGeeks” and their first non-repeating characters are ‘G’, ‘e’ and ‘F’ respectively.

Input: str = “xxyyxx”, q[] = {{2, 3}, {3, 4}}
Output:
-1
y

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

Approach: Create a frequency array freq[][] where freq[i][j] stores the frequency of the character in the sub-string str[0…j] whose ASCII value is i. Now, frequency of any character in the sub-string str[i…j] whose ASCII value is x can be calculated as freq[x][j] = freq[x][i – 1].
Now for every query, start traversing the string in the given range i.e. str[L…R] and for every character, if its frequency is 1 then this is the first non-repeating character in the required sub-string. If all the characters have frequency greater than 1 then print -1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Maximum distinct characters possible ` `#define MAX 256 ` ` `  `// To store the frequency of the characters ` `int` `freq[MAX][MAX]; ` ` `  `// Function to pre-calculate the frequency array ` `void` `preCalculate(string str, ``int` `n) ` `{ ` `    ``// Only the first character has ` `    ``// frequency 1 till index 0 ` `    ``freq[(``int``)str] = 1; ` ` `  `    ``// Starting from the second ` `    ``// character of the string ` `    ``for` `(``int` `i = 1; i < n; i++) ` `    ``{ ` `        ``char` `ch = str[i]; ` ` `  `        ``// For every possible character ` `        ``for` `(``int` `j = 0; j < MAX; j++) ` `        ``{ ` `            ``// Current character under consideration ` `            ``char` `charToUpdate = j; ` ` `  `            ``// If it is equal to the character ` `            ``// at the current index ` `            ``if` `(charToUpdate == ch) ` `                ``freq[j][i] = freq[j][i - 1] + 1; ` `            ``else` `                ``freq[j][i] = freq[j][i - 1]; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to return the frequency of the ` `// given character in the sub-string str[l...r] ` `int` `getFrequency(``char` `ch, ``int` `l, ``int` `r) ` `{ ` `    ``if` `(l == 0) ` `        ``return` `freq[(``int``)ch][r]; ` `    ``else` `        ``return` `(freq[(``int``)ch][r] -  ` `                ``freq[(``int``)ch][l - 1]); ` `} ` ` `  `// Function to return the first ` `// non-repeating character in range[l..r] ` `string firstNonRepeating(string str, ``int` `n,  ` `                              ``int` `l, ``int` `r) ` `{ ` `    ``char` `t = ``""``; ` ` `  `    ``// Starting from the first character ` `    ``for` `(``int` `i = l; i < r; i++) ` `    ``{ ` `        ``// Current character ` `        ``char` `ch = str[i]; ` ` `  `        ``// If frequency of the current character is 1 ` `        ``// then return the character ` `        ``if` `(getFrequency(ch, l, r) == 1) ` `        ``{ ` `            ``t = ch; ` `            ``return` `t; ` `        ``} ` `    ``} ` ` `  `    ``// All the characters of the ` `    ``// sub-string are repeating ` `    ``return` `"-1"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"GeeksForGeeks"``; ` `    ``int` `n = str.length(); ` ` `  `    ``int` `queries[] = {{0, 3}, {2, 3}, {5, 12}}; ` `    ``int` `q = ``sizeof``(queries) /  ` `            ``sizeof``(queries); ` ` `  `    ``// Pre-calculate the frequency array ` `    ``freq[MAX][n] = {0}; ` `    ``preCalculate(str, n); ` ` `  `    ``for` `(``int` `i = 0; i < q; i++) ` `        ``cout << firstNonRepeating(str, n, queries[i],  ` `                                          ``queries[i])  ` `                                                ``<< endl; ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

## Java

 `// Java implementation of the approach ` `public` `class` `GFG { ` ` `  `    ``// Maximum distinct characters possible ` `    ``static` `final` `int` `MAX = ``256``; ` ` `  `    ``// To store the frequency of the characters ` `    ``static` `int` `freq[][]; ` ` `  `    ``// Function to pre-calculate the frequency array ` `    ``static` `void` `preCalculate(String str, ``int` `n) ` `    ``{ ` ` `  `        ``// Only the first character has ` `        ``// frequency 1 till index 0 ` `        ``freq[(``int``)str.charAt(``0``)][``0``] = ``1``; ` ` `  `        ``// Starting from the second ` `        ``// character of the string ` `        ``for` `(``int` `i = ``1``; i < n; i++) { ` `            ``char` `ch = str.charAt(i); ` ` `  `            ``// For every possible character ` `            ``for` `(``int` `j = ``0``; j < MAX; j++) { ` ` `  `                ``// Current character under consideration ` `                ``char` `charToUpdate = (``char``)j; ` ` `  `                ``// If it is equal to the character ` `                ``// at the current index ` `                ``if` `(charToUpdate == ch) ` `                    ``freq[j][i] = freq[j][i - ``1``] + ``1``; ` `                ``else` `                    ``freq[j][i] = freq[j][i - ``1``]; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Function to return the frequency of the ` `    ``// given character in the sub-string str[l...r] ` `    ``static` `int` `getFrequency(``char` `ch, ``int` `l, ``int` `r) ` `    ``{ ` ` `  `        ``if` `(l == ``0``) ` `            ``return` `freq[(``int``)ch][r]; ` `        ``else` `            ``return` `(freq[(``int``)ch][r] - freq[(``int``)ch][l - ``1``]); ` `    ``} ` ` `  `    ``// Function to return the first non-repeating character in range[l..r] ` `    ``static` `String firstNonRepeating(String str, ``int` `n, ``int` `l, ``int` `r) ` `    ``{ ` ` `  `        ``// Starting from the first character ` `        ``for` `(``int` `i = l; i < r; i++) { ` ` `  `            ``// Current character ` `            ``char` `ch = str.charAt(i); ` ` `  `            ``// If frequency of the current character is 1 ` `            ``// then return the character ` `            ``if` `(getFrequency(ch, l, r) == ``1``) ` `                ``return` `(``""` `+ ch); ` `        ``} ` ` `  `        ``// All the characters of the ` `        ``// sub-string are repeating ` `        ``return` `"-1"``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``String str = ``"GeeksForGeeks"``; ` `        ``int` `n = str.length(); ` ` `  `        ``int` `queries[][] = { { ``0``, ``3` `}, { ``2``, ``3` `}, { ``5``, ``12` `} }; ` `        ``int` `q = queries.length; ` ` `  `        ``// Pre-calculate the frequency array ` `        ``freq = ``new` `int``[MAX][n]; ` `        ``preCalculate(str, n); ` ` `  `        ``for` `(``int` `i = ``0``; i < q; i++) { ` `            ``System.out.println(firstNonRepeating(str, n, ` `                                                 ``queries[i][``0``], ` `                                                 ``queries[i][``1``])); ` `        ``} ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Maximum distinct characters possible ` `MAX` `=` `256` ` `  `# To store the frequency of the characters ` `freq ``=` `[[``0` `for` `i ``in` `range``(``MAX``)] ` `           ``for` `j ``in` `range``(``MAX``)] ` ` `  `# Function to pre-calculate  ` `# the frequency array ` `def` `preCalculate(string, n): ` ` `  `    ``# Only the first character has ` `    ``# frequency 1 till index 0 ` `    ``freq[``ord``(string[``0``])][``0``] ``=` `1` ` `  `    ``# Starting from the second ` `    ``# character of the string ` `    ``for` `i ``in` `range``(``1``, n): ` `        ``ch ``=` `string[i] ` ` `  `        ``# For every possible character ` `        ``for` `j ``in` `range``(``MAX``): ` ` `  `            ``# Current character under consideration ` `            ``charToUpdate ``=` `chr``(j) ` ` `  `            ``# If it is equal to the character ` `            ``# at the current index ` `            ``if` `charToUpdate ``=``=` `ch: ` `                ``freq[j][i] ``=` `freq[j][i ``-` `1``] ``+` `1` `            ``else``: ` `                ``freq[j][i] ``=` `freq[j][i ``-` `1``] ` ` `  `# Function to return the frequency of the ` `# given character in the sub-string str[l...r] ` `def` `getFrequency(ch, l, r): ` `    ``if` `l ``=``=` `0``: ` `        ``return` `freq[``ord``(ch)][r] ` `    ``else``: ` `        ``return` `(freq[``ord``(ch)][r] ``-`  `                ``freq[``ord``(ch)][l ``-` `1``]) ` ` `  `# Function to return the first ` `# non-repeating character in range[l..r] ` `def` `firstNonRepeating(string, n, l, r): ` `    ``t ``=` `[""] ``*` `2` ` `  `    ``# Starting from the first character ` `    ``for` `i ``in` `range``(l, r): ` ` `  `        ``# Current character ` `        ``ch ``=` `string[i] ` ` `  `        ``# If frequency of the current character is 1 ` `        ``# then return the character ` `        ``if` `getFrequency(ch, l, r) ``=``=` `1``: ` `            ``t[``0``] ``=` `ch ` `            ``return` `t[``0``] ` ` `  `    ``# All the characters of the ` `    ``# sub-string are repeating ` `    ``return` `"-1"` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``string ``=` `"GeeksForGeeks"` `    ``n ``=` `len``(string) ` ` `  `    ``queries ``=` `[(``0``, ``3``), (``2``, ``3``), (``5``, ``12``)] ` `    ``q ``=` `len``(queries) ` ` `  `    ``# Pre-calculate the frequency array ` `    ``preCalculate(string, n) ` ` `  `    ``for` `i ``in` `range``(q): ` `        ``print``(firstNonRepeating(string, n,  ` `                                ``queries[i][``0``],  ` `                                ``queries[i][``1``])) ` ` `  `# This code is conributed by ` `# sanjeev2552 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Maximum distinct characters possible  ` `    ``static` `int` `MAX = 256;  ` ` `  `    ``// To store the frequency of the characters  ` `    ``static` `int` `[,]freq ;  ` ` `  `    ``// Function to pre-calculate the frequency array  ` `    ``static` `void` `preCalculate(``string` `str, ``int` `n)  ` `    ``{  ` ` `  `        ``// Only the first character has  ` `        ``// frequency 1 till index 0  ` `        ``freq[(``int``)str,0] = 1;  ` ` `  `        ``// Starting from the second  ` `        ``// character of the string  ` `        ``for` `(``int` `i = 1; i < n; i++)  ` `        ``{  ` `            ``char` `ch = str[i];  ` ` `  `            ``// For every possible character  ` `            ``for` `(``int` `j = 0; j < MAX; j++)  ` `            ``{  ` ` `  `                ``// Current character under consideration  ` `                ``char` `charToUpdate = (``char``)j;  ` ` `  `                ``// If it is equal to the character  ` `                ``// at the current index  ` `                ``if` `(charToUpdate == ch)  ` `                    ``freq[j,i] = freq[j,i - 1] + 1;  ` `                ``else` `                    ``freq[j,i] = freq[j,i - 1];  ` `            ``}  ` `        ``}  ` `    ``}  ` ` `  `    ``// Function to return the frequency of the  ` `    ``// given character in the sub-string str[l...r]  ` `    ``static` `int` `getFrequency(``char` `ch, ``int` `l, ``int` `r)  ` `    ``{  ` ` `  `        ``if` `(l == 0)  ` `            ``return` `freq[(``int``)ch, r];  ` `        ``else` `            ``return` `(freq[(``int``)ch, r] - freq[(``int``)ch, l - 1]);  ` `    ``}  ` ` `  `    ``// Function to return the first non-repeating character in range[l..r]  ` `    ``static` `string` `firstNonRepeating(``string` `str, ``int` `n, ``int` `l, ``int` `r)  ` `    ``{  ` ` `  `        ``// Starting from the first character  ` `        ``for` `(``int` `i = l; i < r; i++)  ` `        ``{  ` ` `  `            ``// Current character  ` `            ``char` `ch = str[i];  ` ` `  `            ``// If frequency of the current character is 1  ` `            ``// then return the character  ` `            ``if` `(getFrequency(ch, l, r) == 1)  ` `                ``return` `(``""` `+ ch);  ` `        ``}  ` ` `  `        ``// All the characters of the  ` `        ``// sub-string are repeating  ` `        ``return` `"-1"``;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``string` `str = ``"GeeksForGeeks"``;  ` `        ``int` `n = str.Length;  ` ` `  `        ``int` `[,]queries = { { 0, 3 }, { 2, 3 }, { 5, 12 } };  ` `        ``int` `q = queries.Length;  ` ` `  `        ``// Pre-calculate the frequency array  ` `        ``freq = ``new` `int``[MAX,n];  ` `        ``preCalculate(str, n);  ` ` `  `        ``for` `(``int` `i = 0; i < q; i++) ` `        ``{  ` `            ``Console.WriteLine(firstNonRepeating(str, n,  ` `                                                ``queries[i,0],  ` `                                                ``queries[i,1]));  ` `        ``}  ` `    ``}  ` ` `  `}  ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```G
e
F
```

My Personal Notes arrow_drop_up Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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Improved By : AnkitRai01, sanjeev2552