# Sort a string according to the order defined by another string

• Difficulty Level : Easy
• Last Updated : 23 Jun, 2022

Given two strings (of lowercase letters), a pattern, and a string. The task is to sort strings according to the order defined by the pattern. It may be assumed that the pattern has all characters of the string and all characters in the pattern appear only once.
Examples:

```Input  : pat = "bca", str = "abc"
Output : str = "bca"

Input  : pat = "bxyzca", str = "abcabcabc"
Output : str = "bbbcccaaa"

Input  : pat = "wcyuogmlrdfphitxjakqvzbnes", str = "jcdokai"
Output : str = "codijak"```

Approach 1:
The idea is to first count occurrences of all characters in str and store these counts in a count array. Then traverse pattern from left to right, and for each character pat[i], see how many times it appears in count array and copy this character these many times to str.
Below is the implementation of the above idea.

Implementation:

## C++

 `// C++ program to sort a string according to the``// order defined by a pattern string``#include ``using` `namespace` `std;``const` `int` `MAX_CHAR = 26;` `// Sort str according to the order defined by pattern.``void` `sortByPattern(string& str, string pat)``{``    ``// Create a count array store count of characters in str.``    ``int` `count[MAX_CHAR] = { 0 };` `    ``// Count number of occurrences of each character``    ``// in str.``    ``for` `(``int` `i = 0; i < str.length(); i++)``        ``count[str[i] - ``'a'``]++;` `    ``// Traverse the pattern and print every characters``    ``// same number of times as it appears in str. This``    ``// loop takes O(m + n) time where m is length of``    ``// pattern and n is length of str.``    ``int` `index = 0;``    ``for` `(``int` `i = 0; i < pat.length(); i++)``        ``for` `(``int` `j = 0; j < count[pat[i] - ``'a'``]; j++)``            ``str[index++] = pat[i];``}` `// Driver code``int` `main()``{``    ``string pat = ``"bca"``;``    ``string str = ``"abc"``;``    ``sortByPattern(str, pat);``    ``cout << str;``    ``return` `0;``}`

## Java

 `// Java program to sort a string according to the``// order defined by a pattern string` `class` `GFG {` `    ``static` `int` `MAX_CHAR = ``26``;` `    ``// Sort str according to the order defined by pattern.``    ``static` `void` `sortByPattern(``char``[] str, ``char``[] pat)``    ``{``        ``// Create a count array stor``        ``// count of characters in str.``        ``int` `count[] = ``new` `int``[MAX_CHAR];` `        ``// Count number of occurrences of``        ``// each character in str.``        ``for` `(``int` `i = ``0``; i < str.length; i++) {``            ``count[str[i] - ``'a'``]++;``        ``}` `        ``// Traverse the pattern and print every characters``        ``// same number of times as it appears in str. This``        ``// loop takes O(m + n) time where m is length of``        ``// pattern and n is length of str.``        ``int` `index = ``0``;``        ``for` `(``int` `i = ``0``; i < pat.length; i++) {``            ``for` `(``int` `j = ``0``; j < count[pat[i] - ``'a'``]; j++) {``                ``str[index++] = pat[i];``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``char``[] pat = ``"bca"``.toCharArray();``        ``char``[] str = ``"abc"``.toCharArray();``        ``sortByPattern(str, pat);``        ``System.out.println(String.valueOf(str));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python3 program to sort a string according to``# the order defined by a pattern string``MAX_CHAR ``=` `26` `# Sort str according to the order defined by pattern.``def` `sortByPattern(``str``, pat):``    ` `    ``global` `MAX_CHAR``    ` `    ``# Create a count array store count``    ``# of characters in str.``    ``count ``=` `[``0``] ``*` `MAX_CHAR``    ` `    ``# Count number of occurrences of``    ``# each character in str.``    ``for` `i ``in` `range` `(``0``, ``len``(``str``)):``        ``count[``ord``(``str``[i]) ``-` `97``] ``+``=` `1``    ` `    ``# Traverse the pattern and print every characters``    ``# same number of times as it appears in str. This``    ``# loop takes O(m + n) time where m is length of``    ``# pattern and n is length of str.``    ``index ``=` `0``;``    ``str` `=` `""``    ` `    ``for` `i ``in` `range` `(``0``, ``len``(pat)):``        ``j ``=` `0``        ``while``(j < count[``ord``(pat[i]) ``-` `ord``(``'a'``)]):``            ``str` `+``=` `pat[i]``            ``j ``=` `j ``+` `1``            ``index ``+``=` `1``    ` `    ``return` `str` `# Driver code``pat ``=` `"bca"``str` `=` `"abc"``print``(sortByPattern(``str``, pat))` `# This code is contributed by ihritik`

## C#

 `// C# program to sort a string according to the``// order defined by a pattern string``using` `System;` `class` `GFG {` `    ``static` `int` `MAX_CHAR = 26;` `    ``// Sort str according to the order defined by pattern.``    ``static` `void` `sortByPattern(``char``[] str, ``char``[] pat)``    ``{``        ``// Create a count array stor``        ``// count of characters in str.``        ``int``[] count = ``new` `int``[MAX_CHAR];` `        ``// Count number of occurrences of``        ``// each character in str.``        ``for` `(``int` `i = 0; i < str.Length; i++) {``            ``count[str[i] - ``'a'``]++;``        ``}` `        ``// Traverse the pattern and print every characters``        ``// same number of times as it appears in str. This``        ``// loop takes O(m + n) time where m is length of``        ``// pattern and n is length of str.``        ``int` `index = 0;``        ``for` `(``int` `i = 0; i < pat.Length; i++) {``            ``for` `(``int` `j = 0; j < count[pat[i] - ``'a'``]; j++) {``                ``str[index++] = pat[i];``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``char``[] pat = ``"bca"``.ToCharArray();``        ``char``[] str = ``"abc"``.ToCharArray();``        ``sortByPattern(str, pat);``        ``Console.WriteLine(String.Join(``""``, str));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`bca`

Time complexity: O(m+n) where m is the length of the pattern and n is the length of str.

Auxiliary Space:  O(1)

Approach 2: Using STL

We can pass a comparator to the sort() function in C++ and sort the string according to the pattern.

## C++

 `#include ``using` `namespace` `std;` `// Declaring a vector globally that stores which character``// is occurring first``vector<``int``> position(26, -1);` `//Comparator function``bool` `cmp(``char``& char1, ``char``& char2)``{``    ``return` `position[char1 - ``'a'``] < position[char2 - ``'a'``];``}` `int` `main()``{` `    ``// Pattern``    ``string pat = ``"wcyuogmlrdfphitxjakqvzbnes"``;` `    ``for` `(``int` `i = 0; i < pat.length(); i++) {``        ``if` `(position[pat[i] - ``'a'``] == -1)``            ``position[pat[i] - ``'a'``] = i;``    ``}` `    ``// String to be sorted``    ``string str = ``"jcdokai"``;` `    ``// Passing a comparator to sort function``    ``sort(str.begin(), str.end(), cmp);``    ``cout << str;``}`

## Javascript

 ``

Output

`codijak`

Time complexity: O(m+nlogn) where m is the length of the pattern and n is the length of str.

Auxiliary Space:  O(1)

Exercise: In the above solution, it is assumed that the pattern has all characters of str. Consider a modified version where the pattern may not have all characters and the task is to put all remaining characters (in the string but not in the pattern) at the end. The remaining characters need to be put in alphabetically sorted order.Hint: In the second loop, when increasing the index and putting the character in str, we can also decrease the count at that time. And finally, we traverse the count array to put the remaining characters in alphabetically sorted order.