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Sort the array of strings according to alphabetical order defined by another string
  • Difficulty Level : Easy
  • Last Updated : 12 Mar, 2021

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.

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 compairing 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++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Map to store the characters with their order
// in the new alphabetical order
unordered_map<char, int> h;
// Function that returns true if x < y
// according to the new alphabetical order
bool compare(string x, string y)
    for (int i = 0; i < min(x.size(), y.size()); i++) {
        if (h[x[i]] == h[y[i]])
        return h[x[i]] < h[y[i]];
    return x.size() < y.size();
// 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
    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 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>()
        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)) ==
                else if(str.indexOf(a.charAt(i)) >
                    return 1;
                    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 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)
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)

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