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

01 Jul, 2022
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.  
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.
Initialize 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++
Java
Python3
C#
Javascript
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std; // Function to check if all the// string characters are uniquebool check(string s){     set<char> a;     // Check for repetition in    // characters    for (auto i : s) {        if (a.count(i))            return false;        a.insert(i);    }     return true;} // Function to generate all possible strings// from the given arrayvector<string> helper(vector<string>& arr,                    int ind){     // Base case    if (ind == arr.size())        return { "" };     // Consider every string as    // a starting substring and    // store the generated string    vector<string> tmp        = helper(arr, ind + 1);     vector<string> 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 stringint maxLength(vector<string>& arr){    vector<string> 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 Codeint main(){    vector<string> s;    s.push_back("abcdefgh");     cout << maxLength(s);     return 0;}
// Java program to implement // the above approachimport java.util.*;import java.lang.*; class GFG{ // Function to check if all the// string characters are uniquestatic boolean check(String s){    HashSet<Character> a = new HashSet<>();         // Check for repetition in    // characters    for(int i = 0; i < s.length(); i++)    {        if (a.contains(s.charAt(i)))        {            return false;        }        a.add(s.charAt(i));    }    return true;} // Function to generate all possible//  strings from the given arraystatic ArrayList<String> helper(ArrayList<String> arr,                                int ind){    ArrayList<String> fin = new ArrayList<>();    fin.add("");           // Base case    if (ind == arr.size() )        return fin;         // Consider every string as    // a starting substring and    // store the generated string    ArrayList<String> tmp = helper(arr, ind + 1);         ArrayList<String> 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.size(); i++)    {        String test = tmp.get(i) +                      arr.get(ind);                                 if (check(test))            ret.add(test);    }    return ret;} // Function to find the maximum// possible length of a stringstatic int maxLength(ArrayList<String> arr){    ArrayList<String> tmp = helper(arr, 0);         int len = 0;         // Return max length possible    for(int i = 0; i < tmp.size(); i++)    {        len = len > tmp.get(i).length() ? len :                     tmp.get(i).length();    }           // Return the answer    return len;} // Driver codepublic static void main (String[] args){    ArrayList<String> s = new ArrayList<>();    s.add("abcdefgh");         System.out.println(maxLength(s));}} // This code is contributed by offbeat
# Python3 program to implement# the above approach  # Function to check if all the# string characters are uniquedef check(s):         a = set()      # Check for repetition in    # characters    for i in s:        if i in a:            return False                     a.add(i)      return True  # Function to generate all possible# strings from the given arraydef helper(arr, ind):      # Base case    if (ind == len(arr)):        return [""]      # Consider every string as    # a starting substring and    # store the generated string    tmp = helper(arr, ind + 1)      ret = tmp      # Add current string to result of    # other strings and check if    # characters are unique or not    for i in tmp:        test = i + arr[ind]                 if (check(test)):            ret.append(test)      return ret     # Function to find the maximum# possible length of a stringdef maxLength(arr):     tmp = helper(arr, 0)      l = 0      # Return max length possible    for i in tmp:        l = l if l > len(i) else len(i)      # Return the answer    return l # Driver Codeif __name__=='__main__':         s = []    s.append("abcdefgh")      print(maxLength(s))  # This code is contributed by pratham76
// C# program to implement// the above approachusing System;using System.Collections;using System.Collections.Generic;using System.Text; class GFG{     // Function to check if all the// string characters are uniquestatic bool check(string s){     HashSet<char> a = new HashSet<char>();     // Check for repetition in    // characters    for(int i = 0; i < s.Length; i++)    {        if (a.Contains(s[i]))        {            return false;        }        a.Add(s[i]);    }    return true;} // Function to generate all possible// strings from the given arraystatic 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 stringstatic 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 Codepublic static void Main(string[] args){    ArrayList s = new ArrayList();    s.Add("abcdefgh");     Console.Write(maxLength(s));}} // This code is contributed by rutvik_56
<script>    // Javascript program to implement the above approach         // Function to check if all the    // string characters are unique    function check(s)    {        let a = new Set();         // Check for repetition in        // characters        for(let i = 0; i < s.length; i++)        {            if (a.has(s[i]))            {                return false;            }            a.add(s[i]);        }        return true;    }     // Function to generate all possible    //  strings from the given array    function helper(arr, ind)    {        let fin = [];        fin.push("");         // Base case        if (ind == arr.length)            return fin;         // Consider every string as        // a starting substring and        // store the generated string        let tmp = helper(arr, ind + 1);         let ret = tmp;         // Add current string to result of        // other strings and check if        // characters are unique or not        for(let i = 0; i < tmp.length; i++)        {            let test = tmp[i] + arr[ind];             if (check(test))                ret.push(test);        }        return ret;    }     // Function to find the maximum    // possible length of a string    function maxLength(arr)    {        let tmp = helper(arr, 0);         let len = 0;         // Return max length possible        for(let i = 0; i < tmp.length; i++)        {            len = len > tmp[i].length ? len : tmp[i].length;        }         // Return the answer        return len;    }         let s = [];    s.push("abcdefgh");          document.write(maxLength(s)); // This code is contributed by suresh07.</script>
8
Time Complexity: O(2N) Auxiliary Space: O(N * 2N) 
Efficient Approach (Using Dynamic Programming): 
C++
#include <bits/stdc++.h>using namespace std; int maxLength(vector<string>& A){    vector<bitset<26> > dp        = { bitset<26>() }; // auxiliary dp storage    int res = 0; // will store number of unique chars in                 // resultant string    for (auto& s : A) {        bitset<26> a; // used to track unique chars        for (char c : s)            a.set(c - 'a');        int n = a.count();        if (n < s.size())            continue; // duplicate chars in current string         for (int i = dp.size() - 1; i >= 0; --i) {            bitset<26> c = dp[i];            if ((c & a).any())                continue; // if 1 or more char common            dp.push_back(c | a); // valid concatenation            res = max(res, (int)c.count() + n);        }    }    return res;} int main(){    vector<string> v = { "ab", "cd", "ab" };    int ans = maxLength(v);    cout << ans; // resultant answer string : cfbdghzest    return 0;}
10
Time Complexity: O(N^2) Auxiliary Space: O(N * 26) 
rutvik_56
offbeat
khushboogoyal499
pratham76
RahulJain6
sooda367
suresh07
sweetyty
HashSet
subsequence
substring
Arrays
Hash
Mathematical
Recursion
Strings
Arrays
Hash
Strings
Mathematical
Recursion
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