# Maximize length of the String by concatenating characters from an Array of Strings

Given an array of strings arr[], the task is to find the maximum possible length of a string of distinct characters that can be generated by concatenating of the subsequence of the given array.

Examples:

Input: arr[] = {“ab”, “cd”, “ab”}
Output: 4
Explanation:
All possible combinations are {“”, “ab”, “cd”, “abcd”, “cdab”}.
Therefore, maximum length possible is 4.

Input: arr[] = {“abcdefgh”}
Output: 8
Explanation:
All possible combinations are: “”, “abcdefgh”.
Therefore, the maximum length possible is 8.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use Recursion.
Follow the steps below to solve the problem:

• Iterate from left to right and consider every string as a possible starting substring.
• Iniatialize a HashSet to store the distinct characters encountered so far.
• Once a string is selected as starting substring, check for every remaining string, if it only contains characters which have not occurred before. Append this string as a substring to the current string being generated.
• After performing the above steps, print the maximum length of a string that has been generated.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if all the ` `// string characters are unique ` `bool` `check(string s) ` `{ ` ` `  `    ``set<``char``> a; ` ` `  `    ``// Check for repetation in ` `    ``// characters ` `    ``for` `(``auto` `i : s) { ` `        ``if` `(a.count(i)) ` `            ``return` `false``; ` `        ``a.insert(i); ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Funcyion to generate all possible strings ` `// from the given array ` `vector helper(vector& arr, ` `                      ``int` `ind) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(ind == arr.size()) ` `        ``return` `{ ``""` `}; ` ` `  `    ``// Consider every string as ` `    ``// a starting substring and ` `    ``// store the generated string ` `    ``vector tmp ` `        ``= helper(arr, ind + 1); ` ` `  `    ``vector ret(tmp.begin(), ` `                       ``tmp.end()); ` ` `  `    ``// Add current string to result of ` `    ``// other strings and check if ` `    ``// characters are unique or not ` `    ``for` `(``auto` `i : tmp) { ` `        ``string test = i + arr[ind]; ` `        ``if` `(check(test)) ` `            ``ret.push_back(test); ` `    ``} ` ` `  `    ``return` `ret; ` `} ` ` `  `// Function to find the maximum ` `// possible length of a string ` `int` `maxLength(vector& arr) ` `{ ` `    ``vector tmp = helper(arr, 0); ` ` `  `    ``int` `len = 0; ` ` `  `    ``// Return max length possible ` `    ``for` `(``auto` `i : tmp) { ` `        ``len = len > i.size() ` `                  ``? len ` `                  ``: i.size(); ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `len; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``vector s; ` `    ``s.push_back(``"abcdefgh"``); ` ` `  `    ``cout << maxLength(s); ` ` `  `    ``return` `0; ` `} `

## C#

 `// C# program to implement  ` `// the above approach  ` `using` `System;  ` `using` `System.Collections;  ` `using` `System.Collections.Generic;  ` `using` `System.Text;  ` ` `  `class` `GFG{ ` `     `  `// Function to check if all the ` `// string characters are unique ` `static` `bool` `check(``string` `s) ` `{ ` ` `  `    ``HashSet<``char``> a = ``new` `HashSet<``char``>(); ` ` `  `    ``// Check for repetation in ` `    ``// characters ` `    ``for``(``int` `i = 0; i < s.Length; i++) ` `    ``{ ` `        ``if` `(a.Contains(s[i])) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` `        ``a.Add(s[i]); ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Funcyion to generate all possible ` `//  strings from the given array ` `static` `ArrayList helper(ArrayList arr, ` `                        ``int` `ind) ` `{ ` `     `  `    ``// Base case ` `    ``if` `(ind == arr.Count) ` `        ``return` `new` `ArrayList(){``""``}; ` ` `  `    ``// Consider every string as ` `    ``// a starting substring and ` `    ``// store the generated string ` `    ``ArrayList tmp = helper(arr, ind + 1); ` ` `  `    ``ArrayList ret = ``new` `ArrayList(tmp); ` ` `  `    ``// Add current string to result of ` `    ``// other strings and check if ` `    ``// characters are unique or not ` `    ``for``(``int` `i = 0; i < tmp.Count; i++) ` `    ``{ ` `        ``string` `test = (``string``)tmp[i] + ` `                      ``(``string``)arr[ind]; ` `                       `  `        ``if` `(check(test)) ` `            ``ret.Add(test); ` `    ``} ` `    ``return` `ret; ` `} ` ` `  `// Function to find the maximum ` `// possible length of a string ` `static` `int` `maxLength(ArrayList arr) ` `{ ` `    ``ArrayList tmp = helper(arr, 0); ` ` `  `    ``int` `len = 0; ` ` `  `    ``// Return max length possible ` `    ``for``(``int` `i = 0; i < tmp.Count; i++) ` `    ``{ ` `        ``len = len > ((``string``)tmp[i]).Length ? len :  ` `                    ``((``string``)tmp[i]).Length; ` `    ``} ` `     `  `    ``// Return the answer ` `    ``return` `len; ` `} ` `     `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``ArrayList s = ``new` `ArrayList(); ` `    ``s.Add(``"abcdefgh"``); ` ` `  `    ``Console.Write(maxLength(s)); ` `} ` `} ` ` `  `// This code is contributed by rutvik_56 `

Output:

```8
```

Time Complexity: O(2N)
Auxiliary Space: O(N * 2N)

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Improved By : rutvik_56