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Sort a string in increasing order of given priorities
• Difficulty Level : Hard
• Last Updated : 22 Apr, 2021

Given an alphanumeric string S of length N, the task is to sort the string in increasing order of their priority based on the following conditions:

• Characters with even ASCII values have higher priority than characters with odd ASCII values.
• Even digits have higher priority than odd digits.
• Digits have higher priority than characters.
• For characters or digits having the same parity, the priority is in increasing order of their ASCII values.

Examples:

Input: S = “abcd1234”
Output: 1324bdac
Explanation:
The ASCII value of “a” is 97.
The ASCII value of “b” is 98.
The ASCII value of “c” is 99.
The ASCII value of “d” is 100.
Since characters with even ASCII value have higher priority, “b” and “d” are placed first followed by “a” and “c”.
Similarly, even digits have more priority than odd digits. Therefore, they are placed first.
Since the numbers have a higher priority than characters, the sorted string is “1324bdac”

Output: bda132

Approach: The idea is to separate the characters with odd and even ASCII values and also the digits with odd and even parity. Then, join these substrings in the order of their priorities. Follow the steps below to solve the problem:

1. Initialize two variables, say digits and characters, to store the characters and digits separately.
2. Sort the strings digits and characters with respect to the ASCII table.
3. Now, traverse all the characters in characters and append each character with an odd parity to a variable say oddChars and each character with an even parity to another variable say, evenChars.
4. Similarly, for the string digits, separate the odd and even parity digits, say oddDigs and evenDigs.
5. Finally, concatenate the string as oddChars + evenChars + oddDigs + evenDigs.

Below is the implementation of the above approach:

## C++14

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to sort the string s``// based on the given conditions``string sortString(string s)``{``    ``// Stores digits of given string``    ``string digits = ``""``;` `    ``// Stores characters of given string``    ``string character = ``""``;` `    ``// Iterate over characters of the string``    ``for` `(``int` `i = 0; i < s.size(); ++i) {``        ``if` `(s[i] >= ``'0'` `&& s[i] <= ``'9'``)`` ``{``     ``digits += s[i];`` ``}`` ``else`` ``{``     ``character += s[i];`` ``}`` ``}` ` ``// Sort the string of digits`` ``sort(digits.begin(), digits.end());` ` ``// Sort the string of characters`` ``sort(character.begin(), character.end());` ` ``// Stores odd and even characters`` ``string OddChar = ``""``, EvenChar = ``""``;` ` ``// Seperate odd and even digits`` ``for` `(``int` `i = 0; i < digits.length(); ++i) {``     ``if` `((digits[i] - ``'0'``) % 2 == 1) {``         ``OddChar += digits[i];``     ``}``     ``else` `{``         ``EvenChar += digits[i];``     ``}`` ``}` ` ``// Concatenate strings in the order`` ``// odd characters followed by even`` ``OddChar += EvenChar;` ` ``EvenChar = ``""``;` ` ``// Seperate the odd and even chars`` ``for` `(``int` `i = 0; i < character.length();``      ``++i) {` `     ``if` `((character[i] - ``'a'``) % 2 == 1) {``         ``OddChar += character[i];``     ``}``     ``else` `{``         ``EvenChar += character[i];``     ``}`` ``}` ` ``// Final string`` ``OddChar += EvenChar;` ` ``// Return the final string`` ``return` `OddChar;`` ``}` ` ``// Driver Code`` ``int` `main()`` ``{``     ``// Given string``     ``string s = ``"abcd1234"``;` `     ``// Returns the sorted string``     ``cout << sortString(s);` `     ``return` `0;`` ``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `    ``// Function to sort the String s``    ``// based on the given conditions``    ``static` `String sortString(String s)``    ``{``      ` `        ``// Stores digits of given String``        ``String digits = ``""``;` `        ``// Stores characters of given String``        ``String character = ``""``;` `        ``// Iterate over characters of the String``        ``for` `(``int` `i = ``0``; i < s.length(); ++i) {``            ``if` `(s.charAt(i) >= ``'0'` `&& s.charAt(i) <= ``'9'``) {``                ``digits += s.charAt(i);``            ``} ``else` `{``                ``character += s.charAt(i);``            ``}``        ``}` `        ``// Sort the String of digits``        ``digits = sort(digits);``        ``// Sort the String of characters``        ``character = sort(character);` `        ``// Stores odd and even characters``        ``String OddChar = ``""``, EvenChar = ``""``;` `        ``// Seperate odd and even digits``        ``for` `(``int` `i = ``0``; i < digits.length(); ++i) {``            ``if` `((digits.charAt(i) - ``'0'``) % ``2` `== ``1``) {``                ``OddChar += digits.charAt(i);``            ``} ``else` `{``                ``EvenChar += digits.charAt(i);``            ``}``        ``}` `        ``// Concatenate Strings in the order``        ``// odd characters followed by even``        ``OddChar += EvenChar;``        ``EvenChar = ``""``;` `        ``// Seperate the odd and even chars``        ``for` `(``int` `i = ``0``; i < character.length(); ++i) {` `            ``if` `((character.charAt(i) - ``'a'``) % ``2` `== ``1``) {``                ``OddChar += character.charAt(i);``            ``} ``else` `{``                ``EvenChar += character.charAt(i);``            ``}``        ``}` `        ``// Final String``        ``OddChar += EvenChar;` `        ``// Return the final String``        ``return` `OddChar;``    ``}` `    ``static` `String sort(String inputString)``    ``{``      ` `        ``// convert input string to char array``        ``char` `tempArray[] = inputString.toCharArray();` `        ``// sort tempArray``        ``Arrays.sort(tempArray);` `        ``// return new sorted string``        ``return` `new` `String(tempArray);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Given String``        ``String s = ``"abcd1234"``;` `        ``// Returns the sorted String``        ``System.out.print(sortString(s));` `    ``}``}` `// This code is contributed by shikhasingrajput`

## Python3

 `# Python3 program for the above approach` `# Function to sort the string s``# based on the given conditions``def` `sortString(s):``    ` `    ``# Stores digits of given string``    ``digits ``=` `""``    ` `    ``# Stores characters of given string``    ``character ``=` `""``    ` `    ``# Iterate over characters of the string``    ``for` `i ``in` `range``(``len``(s)):``        ``if` `(s[i] >``=` `'0'` `and` `s[i] <``=` `'9'``):``            ``digits ``+``=` `s[i]``        ``else``:``            ``character ``+``=` `s[i]``            ` `    ``# Sort the string of digits``    ``Digits ``=` `list``(digits)``    ``Digits.sort()``    ` `    ``# Sort the string of characters``    ``Character ``=` `list``(character)``    ``Character.sort()``    ` `    ``# Stores odd and even characters``    ``OddChar, EvenChar ``=` `"``", "``"``    ` `    ``# Seperate odd and even digits``    ``for` `i ``in` `range``(``len``(Digits)):``        ``if` `((``ord``(digits[i]) ``-` `ord``(``'0'``)) ``%` `2` `=``=` `1``):``            ``OddChar ``+``=` `Digits[i]``        ``else``:``            ``EvenChar ``+``=` `Digits[i]``            ` `    ``# Concatenate strings in the order``    ``# odd characters followed by even``    ``OddChar ``+``=` `EvenChar``    ``EvenChar ``=` `""``    ` `    ``# Seperate the odd and even chars``    ``for` `i ``in` `range``(``len``(Character)):``        ``if` `((``ord``(Character[i]) ``-` `ord``(``'a'``)) ``%` `2` `=``=` `1``):``            ``OddChar ``+``=` `Character[i]``        ``else``:``            ``EvenChar ``+``=` `Character[i]``            ` `    ``# Final string``    ``OddChar ``+``=` `EvenChar``    ` `    ``# Return the final string``    ``return` `OddChar``    ` `# Driver Code` `# Given string``s ``=` `"abcd1234"` `# Returns the sorted string``print``(sortString(``list``(s)))` `# This code is contributed by divyesh072019`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{` `  ``// Function to sort the string s``  ``// based on the given conditions``  ``static` `string` `sortString(``char``[] s)``  ``{` `    ``// Stores digits of given string``    ``string` `digits = ``""``;` `    ``// Stores characters of given string``    ``string` `character = ``""``;` `    ``// Iterate over characters of the string``    ``for` `(``int` `i = 0; i < s.Length; ++i)``    ``{``      ``if` `(s[i] >= ``'0'` `&& s[i] <= ``'9'``)``      ``{``        ``digits += s[i];``      ``}``      ``else``      ``{``        ``character += s[i];``      ``}``    ``}` `    ``// Sort the string of digits``    ``char``[] Digits = digits.ToCharArray();``    ``Array.Sort(Digits);` `    ``// Sort the string of characters``    ``char``[] Character = character.ToCharArray();``    ``Array.Sort(Character);` `    ``// Stores odd and even characters``    ``string` `OddChar = ``""``, EvenChar = ``""``;` `    ``// Seperate odd and even digits``    ``for` `(``int` `i = 0; i < Digits.Length; ++i)``    ``{``      ``if` `((digits[i] - ``'0'``) % 2 == 1)``      ``{``        ``OddChar += Digits[i];``      ``}``      ``else``      ``{``        ``EvenChar += Digits[i];``      ``}``    ``}` `    ``// Concatenate strings in the order``    ``// odd characters followed by even``    ``OddChar += EvenChar;``    ``EvenChar = ``""``;` `    ``// Seperate the odd and even chars``    ``for` `(``int` `i = 0; i < Character.Length; ++i)``    ``{` `      ``if` `((Character[i] - ``'a'``) % 2 == 1)``      ``{``        ``OddChar += Character[i];``      ``}``      ``else``      ``{``        ``EvenChar += Character[i];``      ``}``    ``}` `    ``// Final string``    ``OddChar += EvenChar;` `    ``// Return the final string``    ``return` `OddChar;``  ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ` `    ``// Given string``    ``string` `s = ``"abcd1234"``;` `    ``// Returns the sorted string``    ``Console.WriteLine(sortString(s.ToCharArray()));``  ``}``}` `// This code is contributed by divyehrabadiya07`

## Javascript

 ``
Output:
`1324bdac`

Time Complexity: O(N)
Auxiliary Space: O(N)

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