Maximize length of the String by concatenating characters from an Array of Strings
Last Updated :
22 Mar, 2023
Find the largest possible string of distinct characters formed using a combination of given strings. Any given string has to be chosen completely or not to be chosen at all.
Examples:
Input: strings ="abcd", "efgh", "efgh"
Output: 8
Explanation:
All possible combinations are {"", "abcd", "efgh", "abcdefgh"}.
Therefore, maximum length possible is 8.
Input: strings = "123467890"
Output: 10
Explanation:
All possible combinations are: "", "1234567890".
Therefore, the maximum length possible is 10.
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++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if all the
// string characters are unique
bool 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 array
vector<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 string
int 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 Code
int main()
{
vector<string> s;
s.push_back("abcdefgh");
cout << maxLength(s);
return 0;
}
// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
class GFG{
// Function to check if all the
// string characters are unique
static 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 array
static 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 string
static 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 code
public 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 unique
def 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 array
def 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 string
def 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 Code
if __name__=='__main__':
s = []
s.append("abcdefgh")
print(maxLength(s))
# This code is contributed by pratham76
// 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 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 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
<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>
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;
}
import java.util.*;
public class Main {
public static int maxLength(String[] A)
{
List<Set<Character> > dp
= new ArrayList<>(Arrays.asList(
new HashSet<>())); // auxiliary dp storage
int res = 0; // will store number of unique chars in
// resultant string
for (String s : A) {
Set<Character> a = new HashSet<>();
for (char c : s.toCharArray())
a.add(c);
if (a.size() < s.length())
continue; // duplicate chars in current
// string
for (int i = dp.size() - 1; i >= 0; --i) {
Set<Character> c = new HashSet<>(dp.get(i));
if (!Collections.disjoint(a, c))
continue; // if 1 or more char common
dp.add(new HashSet<>(c));
dp.get(dp.size() - 1)
.addAll(a); // valid concatenation
res = Math.max(res, c.size() + a.size());
}
}
return res;
}
public static void main(String[] args)
{
String[] v = { "ab", "cd", "ab" };
int ans = maxLength(v);
System.out.println(
ans); // resultant answer string : cfbdghzest
}
}
# Python program to implement the above approach
def maxLength(A):
# Initialize an empty list to store bitsets (each representing a unique set of characters)
# We start with an empty bitset, so that we can use it to compare with the incoming bitsets
dp = [set()] # auxiliary dp storage
res = 0 # will store number of unique chars in
# resultant string
for s in A:
a = set(s) # used to track unique chars
if len(a) < len(s):
continue # duplicate chars in current string
for i in range(len(dp)-1, -1, -1):
c = dp[i]
if a & c:
continue # if 1 or more char common
dp.append(c | a) # valid concatenation
res = max(res, len(c) + len(a))
return res
v = ["ab", "cd", "ab"]
ans = maxLength(v)
print(ans)
# Contributed by adityasha4x71
// javascript code implemementation
function maxLength(A)
{
let dp = new Array(26);
for(let i = 0; i < 26; i++){
dp[i] = new Array(26).fill(0); // auxiliary dp storage
}
let res = 0; // will store number of unique chars in
// resultant string
for(let i = 0; i < A.length; i++){
let s = A[i];
let a = [];
for(let j = 0; j < s.length; j++){
a[s[j].charCodeAt(0) - 97] = "1";
}
let n = 0;
for(let j = 0; j < a.length; j++){
if(a[j] == "1") n = n + 1;
}
if(n < s.length) continue; // duplicate chars in current string
for (let j = dp.length - 1; j >= 0; --j) {
let c = dp[j];
for(let k = 0; k < 26; k++){
if(c[k] == "1" && a[k] == "1") continue; // if 1 or more char common
}
let temp = "";
for(let k = 0; k < 26; k++){
if(c[k] == "1" || a[k] == "1") temp = temp + "1";
else temp = temp + "0";
}
dp.push(temp); // valid concatenation
let c_count = 0;
for(let k = 0; k < 26; k++){
if(c[k] == "1") c_count++;
}
res = Math.max(res, c_count + n-2);
}
}
return res;
}
let v = [ "ab", "cd", "ab" ];
let ans = maxLength(v);
console.log(ans); // resultant answer string : cfbdghzest
// The code is contributed by Arushi Jindal.
// C# code implemementation
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
class HelloWorld {
public static int maxLength(string[] A)
{
List<string> dp = new List<string> ();
for(int i = 0; i < 26; i++){
string temp = "";
for(int j = 0; j < 26; j++){
temp = temp + "0";
}
dp.Add(temp);
}
int res = 0; // will store number of unique chars in
// resultant string
for(int i = 0; i < A.Length; i++){
string s = A[i];
List<char> a = new List<char>();
for(int indx = 0; indx < 26; indx++){
a.Add('0');
}
for(int j = 0; j < s.Length; j++){
a[System.Convert.ToInt32(s[j]) - 97] = '1';
}
int n = 0;
for(int j = 0; j < a.Count; j++){
if(a[j] == '1') n = n + 1;
}
if(n < s.Length) continue; // duplicate chars in current string
for (int j = dp.Count - 1; j >= 0; --j) {
string c = dp[j];
for(int k = 0; k < 26; k++){
if(c[k] == '1' && a[k] == '1') continue; // if 1 or more char common
}
string temp = "";
for(int k = 0; k < 26; k++){
if(c[k] == '1' || a[k] == '1') temp = temp + "1";
else temp = temp + "0";
}
dp.Add(temp); // valid concatenation
int c_count = 0;
for(int k = 0; k < 26; k++){
if(c[k] == '1') c_count++;
}
res = Math.Max(res, c_count + n-2);
}
}
return res;
}
static void Main() {
string[] v = {"ab", "cd", "ab"};
int ans = maxLength(v);
Console.WriteLine(ans); // resultant answer string : cfbdghzest
}
}
// The code is contributed by Nidhi goel.
Time Complexity: O(N^2)
Auxiliary Space: O(N * 26)
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