Find the last non repeating character in string

Given a string str, the task is to find the last non-repeating character in it. 
For example, if the input string is “GeeksForGeeks”, then the output should be ‘r’ and if the input string is “GeeksQuiz” then the output should be ‘z’. if there is no non-repeating character then print -1.
Examples:

Input: str = “GeeksForGeeks” 
Output: r 
‘r’ is the first character from the end which has frequency 1.
Input: str = “aabbcc” 
Output: -1 
All the characters of the given string have frequencies greater than 1. 

Approach: Create a frequency array that will store the frequency of each of the characters of the given string. Once the frequencies have been updated, start traversing the string from the end character by character and for every character, if the frequency of the current character is 1 then this is the last non-repeating character. If all the characters have a frequency greater than 1 then print -1.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Maximum distinct characters possible
const int MAX = 256;

// Function to return the last non-repeating character
static string lastNonRepeating(string str, int n)
{

    // To store the frequency of each of
    // the character of the given string
    int freq[MAX] = {0};

    // Update the frequencies
    for (int i = 0; i < n; i++)
        freq[str.at(i)]++;


    // Starting from the last character
    for (int i = n - 1; i >= 0; i--)
    {

        // Current character
        char ch = str.at(i);

        // If frequency of the current character is 1
        // then return the character
        if (freq[ch] == 1)
        {
            string res;
            res+=ch;
            return res;
        }
    }

    // All the characters of the
    // string are repeating
    return "-1";
}

// Driver code
int main()
{
    string str = "GeeksForGeeks";
    int n = str.size();
    cout<< lastNonRepeating(str, n);
    return 0;
}

// This code has been contributed by 29AjayKumar

Java

// Java implementation of the approach
public class GFG {

    // Maximum distinct characters possible
    static final int MAX = 256;

    // Function to return the last non-repeating character
    static String lastNonRepeating(String str, int n)
    {

        // To store the frequency of each of
        // the character of the given string
        int freq[] = new int[MAX];

        // Update the frequencies
        for (int i = 0; i < n; i++)
            freq[str.charAt(i)]++;

        // Starting from the last character
        for (int i = n - 1; i >= 0; i--) {

            // Current character
            char ch = str.charAt(i);

            // If frequency of the current character is 1
            // then return the character
            if (freq[ch] == 1)
                return ("" + ch);
        }

        // All the characters of the
        // string are repeating
        return "-1";
    }

    // Driver code
    public static void main(String[] args)
    {
        String str = "GeeksForGeeks";
        int n = str.length();
        System.out.println(lastNonRepeating(str, n));
    }
}

Python3

# Python3 implementation of the approach

# Maximum distinct characters possible
MAX = 256;

# Function to return the last non-repeating character
def lastNonRepeating(string, n) :

    # To store the frequency of each of
    # the character of the given string
    freq = [0]*MAX;

    # Update the frequencies
    for i in range(n) :
        freq[ord(string[i])] += 1;

    # Starting from the last character
    for i in range(n-1,-1,-1) :

        # Current character
        ch = string[i];

        # If frequency of the current character is 1
        # then return the character
        if (freq[ord(ch)] == 1) :
            return ("" + ch);


    # All the characters of the
    # string are repeating
    return "-1";


# Driver code
if __name__ == "__main__" :

    string = "GeeksForGeeks";
    n = len(string);

    print(lastNonRepeating(string, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    // Maximum distinct characters possible
    static readonly int MAX = 256;

    // Function to return the last non-repeating character
    static String lastNonRepeating(String str, int n)
    {

        // To store the frequency of each of
        // the character of the given string
        int []freq = new int[MAX];

        // Update the frequencies
        for (int i = 0; i < n; i++)
            freq[str[i]]++;

        // Starting from the last character
        for (int i = n - 1; i >= 0; i--)
        {

            // Current character
            char ch = str[i];

            // If frequency of the current character is 1
            // then return the character
            if (freq[ch] == 1)
                return ("" + ch);
        }

        // All the characters of the
        // string are repeating
        return "-1";
    }

    // Driver code
    public static void Main(String[] args)
    {
        String str = "GeeksForGeeks";
        int n = str.Length;
        Console.WriteLine(lastNonRepeating(str, n));
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

function lastNonRepeating(str, n) {
    // Maximum distinct characters possible
    const MAX = 256;

    // To store the frequency of each of
    // the character of the given string
    const freq = new Array(MAX).fill(0);

    // Update the frequencies
    for (let i = 0; i < n; i++)
        freq[str.charCodeAt(i)]++;

    // Starting from the last character
    for (let i = n - 1; i >= 0; i--) {
        // Current character
        const ch = str.charAt(i);

        // If frequency of the current character is 1
        // then return the character
        if (freq[ch.charCodeAt()] == 1) return ch;
    }

    // All the characters of the
    // string are repeating
    return "-1";
}

const str = "GeeksForGeeks";
const n = str.length;
console.log(lastNonRepeating(str, n));

Output:

r

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(256) ? O(1), no extra space is required, so it is a constant.