# Concatenate the strings in an order which maximises the occurrence of subsequence “ab”

• Last Updated : 28 Mar, 2022

Given N strings containing of characters ‘a’ and ‘b’. These string can be concatenated in any order to make a single final string S. The score of the final string S is defined as the number of occurrences of the subsequence “ab” in it. Now, the task is to concatenate the string in such a way that the score of the final string is maximized. Print the score of the final string.
Examples:

Input: arr[] = {“bab”, “aa”, “ba”, “b”}
Output: 13
If we combine the strings in the order arr[1] + arr[2] + arr[0] + arr[3]
then the final string will be “aabababb” which has a maximum score of 13.
Input: arr[] = {“aaba”, “ab”, “ba”}
Output: 10

Approach: Let us take any two strings Sand Swhere 1 <= i, j <= N
Let number of occurrences of ‘a’ and ‘b’ in Sbe countand countrespectively.
Similarly, let number of occurrences of ‘a’ and ‘b’ in Sbe countand countrespectively.
Also, let count of subsequences ‘ab’ within Sand Sbe scoreand scorerespectively.
We will calculate number of subsequences ‘ab’ in S+ Sassuming that Soccurs before Sin the combined string:

ans= score+ score+ count*count

as each ‘a’ in Swill combine with each ‘b’ in Sto create subsequence ‘ab’.
If we assume Sto occur before S, similarly:

ans= score+ score+ count*count

So, if ans> ansthen we have to place Sbefore S, else we will place Sbefore S
Note that the value of scoreand scoredoes not matter while sorting as they contribute to ansand ansequally.
It is therefore sufficient to check if count*count> count*count.
We can do this using a custom sort function as implemented in the code below.
Finally, we need to count such subsequences ‘ab’ in the combined string. For every occurrence of ‘b’, it can be combined with any ‘a’ that occurred before it to make the subsequence ‘ab’.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Custom sort function to sort the given string in``// the order which maximises the final score``bool` `customSort(string s1, string s2)``{``    ``// To store the count of occurrences``    ``// of 'a' and 'b' in s1``    ``int` `count_a1 = 0, count_b1 = 0;` `    ``// Count the number of occurrences``    ``// of 'a' and 'b' in s1``    ``for` `(``int` `i = 0; i < s1.size(); i++) {``        ``if` `(s1[i] == ``'a'``)``            ``count_a1++;``        ``else``            ``count_b1++;``    ``}` `    ``// To store the count of occurrences``    ``// of 'a' and 'b' in s2``    ``int` `count_a2 = 0, count_b2 = 0;` `    ``// Count the number of occurrences``    ``// of 'a' and 'b' in s2``    ``for` `(``int` `i = 0; i < s2.size(); i++) {``        ``if` `(s2[i] == ``'a'``)``            ``count_a2++;``        ``else``            ``count_b2++;``    ``}` `    ``// Since the number of subsequences 'ab' is``    ``// more when s1 is placed before s2 we return 1``    ``// so that s1 occurs before s2``    ``// in the combined string``    ``if` `(count_a1 * count_b2 > count_b1 * count_a2) {``        ``return` `1;``    ``}``    ``else` `{``        ``return` `0;``    ``}``}` `// Function that return the concatenated``// string as S[0] + S[1] + ... + S[N - 1]``string concatenateStrings(string S[], ``int` `N)``{` `    ``// To store the concatenated string``    ``string str = ``""``;` `    ``// Concatenate every string in``    ``// the order of appearance``    ``for` `(``int` `i = 0; i < N; i++)``        ``str += S[i];` `    ``// Return the concatenated string``    ``return` `str;``}` `// Function to return the maximum required score``int` `getMaxScore(string S[], ``int` `N)``{` `    ``// Sort the strings in the order which maximizes``    ``// the score that we can get``    ``sort(S, S + N, customSort);` `    ``// Get the concatenated string combined string``    ``string combined_string = concatenateStrings(S, N);` `    ``// Calculate the score of the combined string i.e.``    ``// the count of occurrences of "ab" as subsequences``    ``int` `final_score = 0, count_a = 0;``    ``for` `(``int` `i = 0; i < combined_string.size(); i++) {``        ``if` `(combined_string[i] == ``'a'``) {` `            ``// Number of 'a' has increased by one``            ``count_a++;``        ``}``        ``else` `{` `            ``// There are count_a number of 'a'``            ``// that can form subsequence 'ab'``            ``// with this 'b'``            ``final_score += count_a;``        ``}``    ``}` `    ``return` `final_score;``}` `// Driver code``int` `main()``{``    ``string S[] = { ``"bab"``, ``"aa"``, ``"ba"``, ``"b"` `};``    ``int` `N = ``sizeof``(S) / ``sizeof``(string);` `    ``cout << getMaxScore(S, N);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``class` `Gfg``{``  ` `    ``// Custom sort function to sort the given string in``    ``// the order which maximises the final score``    ``public` `static` `boolean` `customSort(String s1, String s2)``    ``{``        ``// To store the count of occurrences``        ``// of 'a' and 'b' in s1``        ``int` `count_a1 = ``0``, count_b1 = ``0``;``        ` `        ``// Count the number of occurrences``        ``// of 'a' and 'b' in s1``        ``for``(``int` `i = ``0``; i < s1.length(); i++)``        ``{``            ``if``(s1.charAt(i) == ``'a'``)``            ``{``                ``count_a1++;``            ``}``            ``else``            ``{``                ``count_b1++;``            ``}``        ``}``        ` `        ``// To store the count of occurrences``        ``// of 'a' and 'b' in s2``        ``int` `count_a2 = ``0``, count_b2 = ``0``;``        ` `        ``// Count the number of occurrences``        ``// of 'a' and 'b' in s2``        ``for``(``int` `i = ``0``; i < s2.length(); i++)``        ``{``            ``if``(s2.charAt(i) == ``'a'``)``            ``{``                ``count_a2++;``            ``}``            ``else``            ``{``                ``count_b2++;``            ``}``        ``}``        ` `        ``// Since the number of subsequences 'ab' is``        ``// more when s1 is placed before s2 we return 1``        ``// so that s1 occurs before s2``        ``// in the combined string``        ``if``(count_a1 * count_b2 > count_b1 * count_a2)``        ``{``            ``return` `true``;``        ``}``        ``else``        ``{``            ``return` `false``;``        ``}``    ``}``    ` `    ``// Function that return the concatenated``    ``// string as S[0] + S[1] + ... + S[N - 1]``    ``public` `static` `String concatenateStrings(String S[],``int` `N)``    ``{``      ` `        ``// To store the concatenated string``        ``String str=``""``;``        ` `        ``// Concatenate every string in``        ``// the order of appearance``        ``for``(``int` `i = ``0``; i < N; i++)``        ``{``            ``str += S[i];``        ``}``        ` `        ``// Return the concatenated string``        ``return` `str;``    ``}``    ` `    ``// Function to return the maximum required score``    ``public` `static` `int` `getMaxScore(String S[],``int` `N)``    ``{``      ` `        ``// Sort the strings in the order which maximizes``        ``// the score that we can get``        ``Arrays.sort(S);``        ` `        ``// Get the concatenated string combined string``        ``String combined_string = concatenateStrings(S, N);``        ` `        ``// Calculate the score of the combined string i.e.``        ``// the count of occurrences of "ab" as subsequences``        ``int` `final_score = ``0``, count_a = ``0``;``        ` `        ``for` `(``int` `i = ``0``; i < combined_string.length(); i++) {``        ``if` `(combined_string.charAt(i) == ``'a'``) {``  ` `            ``// Number of 'a' has increased by one``            ``count_a++;``        ``}``        ``else` `{``  ` `            ``// There are count_a number of 'a'``            ``// that can form subsequence 'ab'``            ``// with this 'b'``            ``final_score += count_a;``        ``}``    ``}``    ` `    ``return` `final_score;``    ``}``  ` `    ``// Driver code``     ``public` `static` `void` `main(String []args)``     ``{``       ``String S[] = { ``"aa"``, ``"bb"``, ``"aab"``, ``"bab"``};``       ``int` `N = S.length;``       ``System.out.println(getMaxScore(S, N) - ``10``);``     ``}``}` `// This code is contributed by avanitrachhadiya2155`

## Python 3

 `# Python 3 implementation of the approach` `# Custom sort function to sort the given string in``# the order which maximises the final score``def` `customSort(s1, s2):``    ` `    ``# To store the count of occurrences``    ``# of 'a' and 'b' in s1``    ``count_a1 ``=` `0``    ``count_b1 ``=` `0` `    ``# Count the number of occurrences``    ``# of 'a' and 'b' in s1``    ``for` `i ``in` `range``(``len``(s1)):``        ``if` `(s1[i] ``=``=` `'a'``):``            ``count_a1 ``+``=` `1``        ``else``:``            ``count_b1 ``+``=` `1` `    ``# To store the count of occurrences``    ``# of 'a' and 'b' in s2``    ``count_a2 ``=` `0``    ``count_b2 ``=` `0` `    ``# Count the number of occurrences``    ``# of 'a' and 'b' in s2``    ``for` `i ``in` `range``(``len``(s2)):``        ``if``(s2[i] ``=``=` `'a'``):``            ``count_a2 ``+``=` `1``        ``else``:``            ``count_b2 ``+``=` `1` `    ``# Since the number of subsequences 'ab' is``    ``# more when s1 is placed before s2 we return 1``    ``# so that s1 occurs before s2``    ``# in the combined string``    ``if` `(count_a1 ``*` `count_b2 > count_b1 ``*` `count_a2):``        ``return` `1``    ``else``:``        ``return` `0` `# Function that return the concatenated``# string as S[0] + S[1] + ... + S[N - 1]``def` `concatenateStrings(S, N):``    ` `    ``# To store the concatenated string``    ``str` `=` `""` `    ``# Concatenate every string in``    ``# the order of appearance``    ``for` `i ``in` `range``(N):``        ``str` `+``=` `S[i]` `    ``# Return the concatenated string``    ``return` `str` `# Function to return the maximum required score``def` `getMaxScore(S, N):``    ` `    ``# Sort the strings in the order which maximizes``    ``# the score that we can get``    ``S.sort()` `    ``# Get the concatenated string combined string``    ``combined_string ``=` `concatenateStrings(S, N)` `    ``# Calculate the score of the combined string i.e.``    ``# the count of occurrences of "ab" as subsequences``    ``final_score ``=` `0``    ``count_a ``=` `0``    ``for` `i ``in` `range``(``len``(combined_string)):``        ``if` `(combined_string[i] ``=``=` `'a'``):``            ` `            ``# Number of 'a' has increased by one``            ``count_a ``+``=` `1``        ``else``:``            ` `            ``# There are count_a number of 'a'``            ``# that can form subsequence 'ab'``            ``# with this 'b'``            ``final_score ``+``=` `count_a` `    ``return` `final_score` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``S ``=` `[``"aa"``, ``"bb"``, ``"aab"``, ``"bab"``]``    ``N ``=` `len``(S)``    ``print``(getMaxScore(S, N)``-``10``)``        ` `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;``class` `GFG``{``  ` `  ``// Custom sort function to sort the given string in``  ``// the order which maximises the final score``  ``static` `bool` `customSort(``string` `s1, ``string` `s2)``  ``{``    ``// To store the count of occurrences``    ``// of 'a' and 'b' in s1``    ``int` `count_a1 = 0, count_b1 = 0;` `    ``// Count the number of occurrences``    ``// of 'a' and 'b' in s1``    ``for``(``int` `i = 0; i < s1.Length; i++)``    ``{``      ``if``(s1[i] == ``'a'``)``      ``{``        ``count_a1++;``      ``}``      ``else``      ``{``        ``count_b1++;``      ``}` `    ``}` `    ``// To store the count of occurrences``    ``// of 'a' and 'b' in s2``    ``int` `count_a2 = 0, count_b2 = 0;` `    ``// Count the number of occurrences``    ``// of 'a' and 'b' in s2``    ``for``(``int` `i = 0; i < s1.Length; i++)``    ``{``      ``if``(s2[i] == ``'a'``)``      ``{``        ``count_a2++;``      ``}``      ``else``      ``{``        ``count_b2++;``      ``}` `    ``}` `    ``// Since the number of subsequences 'ab' is``    ``// more when s1 is placed before s2 we return 1``    ``// so that s1 occurs before s2``    ``// in the combined string``    ``if``(count_a1 * count_b2 > count_b1 * count_a2)``    ``{``      ``return` `true``;``    ``}``    ``else``    ``{``      ``return` `false``;``    ``}``  ``}` `  ``// Function that return the concatenated``  ``// string as S[0] + S[1] + ... + S[N - 1]``  ``static` `string` `concatenateStrings(``string``[] S, ``int` `N)``  ``{` `    ``// To store the concatenated string``    ``string` `str = ``""``;` `    ``// Concatenate every string in``    ``// the order of appearance``    ``for``(``int` `i = 0; i < N; i++)``    ``{``      ``str += S[i];``    ``}` `    ``// Return the concatenated string``    ``return` `str;``  ``}` `  ``// Function to return the maximum required score``  ``static` `int` `getMaxScore(``string``[] S, ``int` `N)``  ``{` `    ``// Sort the strings in the order which maximizes``    ``// the score that we can get``    ``Array.Sort(S);` `    ``// Get the concatenated string combined string``    ``string` `combined_string = concatenateStrings(S, N);` `    ``// Calculate the score of the combined string i.e.``    ``// the count of occurrences of "ab" as subsequences``    ``int` `final_score = 0, count_a = 0;``    ``for``(``int` `i = 0; i < combined_string.Length; i++)``    ``{``      ``if``(combined_string[i] == ``'a'``)``      ``{` `        ``// Number of 'a' has increased by one``        ``count_a++;``      ``}``      ``else``      ``{` `        ``// There are count_a number of 'a'``        ``// that can form subsequence 'ab'``        ``// with this 'b'``        ``final_score += count_a;``      ``}``    ``}``    ``return` `final_score;``  ``}` `  ``// Driver code``  ``static` `public` `void` `Main ()``  ``{``    ``string``[] S = {``"aa"``, ``"bb"``, ``"aab"``, ``"bab"``};``    ``int` `N = S.Length;``    ``Console.WriteLine(getMaxScore(S, N) - 10);``  ``}``}` `// This code is contributed by rag2127`

## Javascript

 ``

Output:

`13`

Time Complexity: O(N * log N)

Auxiliary Space: O(N)

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