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

28 Feb, 2022
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 theInput: 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++
Java
Python3
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Map to store the characters with their order// in the new alphabetical orderunordered_map<char, int> h; // Function that returns true if x < y// according to the new alphabetical orderbool compare(string x, string y){    for (int i = 0; i < min(x.size(), y.size()); i++) {        if (h[x[i]] == h[y[i]])            continue;        return h[x[i]] < h[y[i]];    }    return x.size() < y.size();} // Driver codeint 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.size(); i++)        h[str[i]] = i;        sort(v.begin(), v.end(), compare);     // Print the strings after sorting    for (auto x : v)        cout << x << " ";         return 0;}
// Java implementation of the approachimport 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 Codepublic 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 implementation of the approach # Function to sort and print the array# according to the new alphabetical orderdef 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 codes = "fguecbdavwyxzhijklmnopqrst"a = ["geeksforgeeks", "is", "the", "best", "place", "for", "learning"]n = len(a)sortStringArray(s, a, n)
<script> // JavaScript implementation of the approach // Map to store the characters with their order// in the new alphabetical orderlet h = new Map(); // Function that returns true if x < y// according to the new alphabetical orderfunction 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 codelet str = "fguecbdavwyxzhijklmnopqrst";let v = [ "geeksforgeeks", "is", "the","best", "place", "for", "learning" ]; // Store the order for each character// in the new alphabetical sequenceh.clear();for (let i = 0; i < str.length; i++)    h.set(str[i],i); v.sort(compare); // Print the strings after sortingfor (let x of v)    document.write(x + " "); // This code is contributed by shinjanpatra</script>
for geeksforgeeks best is learning place the
 
Time Complexity: O(N * log(N)), where N is the size of the string str
Auxiliary Space: O(N)
dungeondragon22
subhammahato348
simranarora5sos
shinjanpatra
Advanced Data Structure
Algorithms
Competitive Programming
Data Structures
Hash
Sorting
Data Structures
Hash
Sorting
Algorithms
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Agents in Artificial Intelligence
Decision Tree Introduction with example
Segment Tree | Set 1 (Sum of given range)
Binary Indexed Tree or Fenwick Tree
AVL Tree | Set 2 (Deletion)
DSA Sheet by Love Babbar
SDE SHEET - A Complete Guide for SDE Preparation
Top 50 Array Coding Problems for Interviews
Difference between BFS and DFS