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

Last Updated : 25 Apr, 2023

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 store         // 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 store         // 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:

We can pass a comparator to the sort() function 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 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; }

## Java

 import java.util.*;   class Main {       // Declaring a list globally that stores which character is occurring first     static List position = new ArrayList<>(Collections.nCopies(26, -1));       // Comparator function     static int cmp(char char1, char char2) {         if (position.get(char1 - 'a') < position.get(char2 - 'a')) {             return -1;         } else if (position.get(char1 - 'a') > position.get(char2 - 'a')) {             return 1;         } else {             return 0;         }     }       public static void main(String[] args) {         // Pattern         String pat = "wcyuogmlrdfphitxjakqvzbnes";           for (int i = 0; i < pat.length(); i++) {             if (position.get(pat.charAt(i) - 'a') == -1) {                 position.set(pat.charAt(i) - 'a', i);             }         }           // String to be sorted         String str = "jcdokai";           // Passing a comparator to the sorted function         char[] charArr = str.toCharArray();         Arrays.sort(charArr, new Comparator() {             public int compare(Character c1, Character c2) {                 return cmp(c1, c2);             }         });         String sortedStr = new String(charArr);         System.out.println(sortedStr);     } }

## Python3

 from typing import List from functools import cmp_to_key   # Declaring a list globally that stores which character is occurring first position: List[int] = [-1] * 26   # Comparator function def cmp(char1: str, char2: str) -> int:     if position[ord(char1) - ord('a')] < position[ord(char2) - ord('a')]:         return -1     elif position[ord(char1) - ord('a')] > position[ord(char2) - ord('a')]:         return 1     else:         return 0   if __name__ == '__main__':     # Pattern     pat = "wcyuogmlrdfphitxjakqvzbnes"       for i in range(len(pat)):         if position[ord(pat[i]) - ord('a')] == -1:             position[ord(pat[i]) - ord('a')] = i       # String to be sorted     str = "jcdokai"       # Passing a comparator to the sorted function     sorted_str = sorted(str, key=cmp_to_key(cmp))     print(''.join(sorted_str))   # This code is contributed by adityashatmfh



## C#

 // C# program for the above approach   using System; using System.Collections.Generic; using System.Linq;   class GFG {       // Declaring a list globally that stores which character is occurring first     static List position = Enumerable.Repeat(-1, 26).ToList();       // Comparator function     static int Cmp(char char1, char char2) {         if (position[char1 - 'a'] < position[char2 - 'a']) {             return -1;         } else if (position[char1 - 'a'] > position[char2 - 'a']) {             return 1;         } else {             return 0;         }     }       public static void 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 the sorted function         char[] charArr = str.ToCharArray();         Array.Sort(charArr, new Comparison(Cmp));         string sortedStr = new string(charArr);         Console.WriteLine(sortedStr);     } }   // This code is contributed by sdeadityasharma

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.