Sort a string according to the order defined by another string

06 Jul, 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++
Java
Python3
C#
Javascript
// C++ program to sort a string according to the// order defined by a pattern string#include <bits/stdc++.h>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 codeint main(){    string pat = "bca";    string str = "abc";    sortByPattern(str, pat);    cout << str;    return 0;}
// 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 program to sort a string according to# the order defined by a pattern stringMAX_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 codepat = "bca"str = "abc"print(sortByPattern(str, pat)) # This code is contributed by ihritik
// C# program to sort a string according to the// order defined by a pattern stringusing 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 */
<script>// Javascript program to sort a string according to the// order defined by a pattern stringlet MAX_CHAR = 26; // Sort str according to the order defined by pattern.function sortByPattern(str,pat){    // Create a count array stor        // count of characters in str.        let count = new Array(MAX_CHAR);        for(let i = 0; i < MAX_CHAR; i++)        {            count[i] = 0;        }           // Count number of occurrences of        // each character in str.        for (let i = 0; i < str.length; i++) {            count[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;        }           // 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.        let index = 0;        for (let i = 0; i < pat.length; i++) {            for (let j = 0; j < count[pat[i].charCodeAt(0) - 'a'.charCodeAt(0)]; j++) {                str[index++] = pat[i];            }        }} // Driver codelet pat = "bca".split("");let str = "abc".split("");sortByPattern(str, pat);document.write((str).join("")); // This code is contributed by rag2127</script>
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++
Javascript
#include <bits/stdc++.h>using namespace std; // Declaring a vector globally that stores which character// is occurring firstvector<int> position(26, -1); //Comparator functionbool 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;}
<script> // Declaring a vector globally that stores which character// is occurring firstlet position = new Array(26).fill(-1); //Comparator functionfunction cmp(char1, char2){    return position[char1.charCodeAt(0) - 'a'.charCodeAt(0)] - position[char2.charCodeAt(0) - 'a'.charCodeAt(0)];} // driver code // Patternlet pat = "wcyuogmlrdfphitxjakqvzbnes"; for (let i = 0; i <br pat.length; i++) {    if (position[pat.charCodeAt(i) - 'a'.charCodeAt(0)] == -1)        position[pat.charCodeAt(i) - 'a'.charCodeAt(0)] = i;} // String to be sortedlet str = "jcdokai"; // Passing a comparator to sort functionstr = str.split("").sort(cmp).join("");document.write(str,"</br>");     // This code is contributed by Shinjan Patra    </script>
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.
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