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< char , int > h;
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();
}
int main()
{
string str = "fguecbdavwyxzhijklmnopqrst" ;
vector<string> v { "geeksforgeeks" , "is" , "the" , "best" , "place" , "for" , "learning" };
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; });
for ( auto x : v)
cout << x << " " ;
return 0;
}
|
C#
using System;
using System.Collections.Generic;
class Program
{
private static Dictionary< char , int > h = new Dictionary< char , int >();
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;
}
static void Main( string [] args)
{
string str = "fguecbdavwyxzhijklmnopqrst" ;
List< string > v = new List< string > { "geeksforgeeks" , "is" , "the" ,
"best" , "place" , "for" , "learning" };
h.Clear();
for ( int i = 0; i < str.Length; i++)
h[str[i]] = i;
v.Sort((x, y) => Compare(x, y));
foreach ( var x in v)
Console.Write(x + " " );
Console.ReadLine();
}
}
|
Java
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);
}
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
def sortStringArray(s, a, n):
a = sorted (a, key = lambda word: [s.index(c) for c in word])
for i in a:
print (i, end = ' ' )
s = "fguecbdavwyxzhijklmnopqrst"
a = [ "geeksforgeeks" , "is" , "the" , "best" , "place" , "for" , "learning" ]
n = len (a)
sortStringArray(s, a, n)
|
Javascript
<script>
let h = new Map();
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;
}
let str = "fguecbdavwyxzhijklmnopqrst" ;
let v = [ "geeksforgeeks" , "is" , "the" , "best" , "place" , "for" , "learning" ];
h.clear();
for (let i = 0; i < str.length; i++)
h.set(str[i],i);
v.sort(compare);
for (let x of v)
document.write(x + " " );
</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)
Last Updated :
23 Feb, 2023
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