Sort the array of strings according to alphabetical order defined by another string

Given a string str and an array of strings strArr[], the task is to sort the array according to the alphabetical order defined by str. 

Note: str and every string in strArr[] consists of only lower case alphabets.

Examples:

Input: str = “fguecbdavwyxzhijklmnopqrst”, 
strArr[] = {“geeksforgeeks”, “is”, “the”, “best”, “place”, “for”, “learning”} 
Output: for geeksforgeeks best is learning place the

Input: str = “avdfghiwyxzjkecbmnopqrstul”, 
strArr[] = {“rainbow”, “consists”, “of”, “colours”} 
Output: consists colours of rainbow 

Approach: Traverse every character of str and store the value in a map with character as the key and its index in the array as the value. 
Now, this map will act as the new alphabetical order of the characters. Start comparing the string in the strArr[] and instead of comparing the ASCII values of the characters, compare the values mapped to those particular characters in the map i.e. if character c1 appears before character c2 in str then c1 < c2.

Below is the implementation of the above approach:

C++

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>

using namespace std;

// Map to store the characters with their order
// in the new alphabetical order
map<char, int> h;

// Function that returns true if x < y
// according to the new alphabetical order
int Compare(string x, string y)
{
    int minSize = min(x.length(), y.length());

    for (int i = 0; i < minSize; i++)
    {
        if (h[x[i]] == h[y[i]])
            continue;
        return h[x[i]] - h[y[i]];
    }

    return x.length() - y.length();
}

// Driver code
int main()
{
    string str = "fguecbdavwyxzhijklmnopqrst";
    vector<string> v { "geeksforgeeks", "is", "the", "best", "place", "for", "learning" };

    // Store the order for each character
    // in the new alphabetical sequence
    h.clear();
    for (int i = 0; i < str.length(); i++)
        h[str[i]] = i;

    sort(v.begin(), v.end(), [](string x, string y) { return Compare(x, y) < 0; });

    // Print the strings after sorting
    for (auto x : v)
        cout << x << " ";

    return 0;
}

C#

using System;
using System.Collections.Generic;

class Program
{
    // Map to store the characters with their order
    // in the new alphabetical order
    private static Dictionary<char, int> h = new Dictionary<char, int>();

    // Function that returns true if x < y
    // according to the new alphabetical order
    private static int Compare(string x, string y)
    {
        int minSize = Math.Min(x.Length, y.Length);

        for (int i = 0; i < minSize; i++)
        {
            if (h[x[i]] == h[y[i]])
                continue;
            return h[x[i]] - h[y[i]];
        }

        return x.Length - y.Length;
    }

    // Driver code
    static void Main(string[] args)
    {
        string str = "fguecbdavwyxzhijklmnopqrst";
        List<string> v = new List<string> { "geeksforgeeks", "is", "the",
                                            "best", "place", "for", "learning" };

        // Store the order for each character
        // in the new alphabetical sequence
        h.Clear();
        for (int i = 0; i < str.Length; i++)
            h[str[i]] = i;

        v.Sort((x, y) => Compare(x, y));

        // Print the strings after sorting
        foreach (var x in v)
            Console.Write(x + " ");

        Console.ReadLine();
    }
}

Java

// Java implementation of the approach
import java.util.Arrays;
import java.util.Comparator;

public class GFG
{
private static void sort(String[] strArr, String str)
{
    Comparator<String> myComp = new Comparator<String>()
    {
        @Override
        public int compare(String a, String b)
        {
            for(int i = 0;
                    i < Math.min(a.length(),
                                 b.length()); i++)
            {
                if (str.indexOf(a.charAt(i)) ==
                    str.indexOf(b.charAt(i)))
                {
                    continue;
                }
                else if(str.indexOf(a.charAt(i)) >
                        str.indexOf(b.charAt(i)))
                {
                    return 1;
                }
                else
                {
                    return -1;
                }
            }
            return 0;
        }
    };
    Arrays.sort(strArr, myComp);
}

// Driver Code
public static void main(String[] args)
{
    String str = "fguecbdavwyxzhijklmnopqrst";
    String[] strArr = {"geeksforgeeks", "is", "the", "best",
                       "place", "for", "learning"};

    sort(strArr, str);

    for(int i = 0; i < strArr.length; i++)
    {
        System.out.print(strArr[i] + " ");
    }
}
}

Python3

# Python3 implementation of the approach

# Function to sort and print the array
# according to the new alphabetical order
def sortStringArray(s, a, n):

    # Sort the array according to the new alphabetical order
    a = sorted(a, key = lambda word: [s.index(c) for c in word])
    for i in a:
        print(i, end =' ')

# Driver code
s = "fguecbdavwyxzhijklmnopqrst"
a = ["geeksforgeeks", "is", "the", "best", "place", "for", "learning"]
n = len(a)
sortStringArray(s, a, n)

Javascript

<script>

// JavaScript implementation of the approach

// Map to store the characters with their order
// in the new alphabetical order
let h = new Map();

// Function that returns true if x < y
// according to the new alphabetical order
function compare(x, y)
{
    for (let i = 0; i < Math.min(x.length, y.length); i++) {
        if (h.get(x[i]) == h.get(y[i]))
            continue;
        return h.get(x[i]) - h.get(y[i]);
    }
    return x.length - y.length;
}

// Driver code
let str = "fguecbdavwyxzhijklmnopqrst";
let v = [ "geeksforgeeks", "is", "the","best", "place", "for", "learning" ];

// Store the order for each character
// in the new alphabetical sequence
h.clear();
for (let i = 0; i < str.length; i++)
    h.set(str[i],i);

v.sort(compare);

// Print the strings after sorting
for (let x of v)
    document.write(x + " ");

// This code is contributed by shinjanpatra
</script>

Output

for geeksforgeeks best is learning place the 

Complexity Analysis:

• Time Complexity: O(N * log(N)), where N is the size of the string str
• Auxiliary Space: O(N)