# Maximum length of string formed by concatenation having even frequency of each character

• Last Updated : 26 Oct, 2021

Given N strings, print the maximum length of the string and the string formed by concatenating any of the N strings, such that every letter in the string occurs even number of times

Example:

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Input: N = 5, str = [“ABAB”, “ABF”, “CDA”, “AD”, “CCC”]
Explanation: The string formed by concatenation is ABABCDAADCCC. Each letter in the string occurs even number of times

Input: N = 3, str = [“AB”, “BC”, “CA”]
Output: ABBCCA 6
Explanation: The string formed by concatenation of all 3 strings is ABBCCA

Approach: The given problem can be solved using recursion and backtracking. The idea is to either include the string or exclude the string at every iteration. After including a string, the frequency of all the characters in the concatenated string is calculated. If frequency of all the characters is even we update the maximum length max. Below steps can be followed to solve the problem:

• Initialize variable max to 0 for calculating maximum length of concatenated string having even frequency of all characters
• Initialize string ans1 to store the concatenated string of maximum length with all character having even frequency
• The base case of the recursive call is to return, if index becomes equal to the size of the input string list
• At every recursive call we perform the following operation:
• Include the string and check if the frequency of characters is even for the concatenated string
• If the frequency is even, update max and ans1
• Increment the index and make the next recursive call
• Exclude the string, increment the index and make the next recursive call

Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach #include using namespace std; int maxi = 0; string ans1 = "";   // Function to check the string void calculate(string ans) {       int dp[26] = { 0 };     for (int i = 0; i < ans.length(); ++i) {           // Count the frequency         // of the string         dp[ans[i] - 'A']++;     }       // Check the frequency of the string     for (int i = 0; i < 26; ++i) {         if (dp[i] % 2 == 1) {             return;         }     }     if (maxi < ans.length()) {           // Store the length         // of the new String         maxi = ans.length();         ans1 = ans;     } }   // Function to find the longest // concatenated string having // every character of even frequency void longestString(vector arr, int index,                    string str) {       // Checking the string     if (index == arr.size()) {         return;     }       // Dont Include the string     longestString(arr, index + 1, str);       // Include the string     str += arr[index];       calculate(str);     longestString(arr, index + 1, str); }   // Driver code int main() {     vector A         = { "ABAB", "ABF", "CDA", "AD", "CCC" };         // Call the function     longestString(A, 0, "");       // Print the answer     cout << ans1 << " " << ans1.length();       return 0; }   // This code is contributed by Potta Lokesh

## Java

 // Java Implementation of the above approach   import java.io.*; import java.util.*;   public class index {     static int max = 0;     static String ans1 = "";       // Function to check the string     static void calculate(String ans)     {           int dp[] = new int[26];         for (int i = 0; i < ans.length(); ++i) {               // Count the frequency             // of the string             dp[ans.charAt(i) - 'A']++;         }           // Check the frequency of the string         for (int i = 0; i < dp.length; ++i) {             if (dp[i] % 2 == 1) {                 return;             }         }         if (max < ans.length()) {               // Store the length             // of the new String             max = ans.length();             ans1 = ans;         }     }       // Function to find the longest     // concatenated string having     // every character of even frequency     static void longestString(         List arr, int index, String str)     {           // Checking the string         if (index == arr.size()) {             return;         }           // Dont Include the string         longestString(arr, index + 1, str);           // Include the string         str += arr.get(index);           calculate(str);         longestString(arr, index + 1, str);     }       // Driver code     public static void main(String[] args)     {         ArrayList A = new ArrayList<>();         A.add("ABAB");         A.add("ABF");         A.add("CDA");         A.add("AD");         A.add("CCC");           // Call the function         longestString(A, 0, "");           // Print the answer         System.out.println(ans1 + " "                            + ans1.length());     } }

## Python3

 # Python3 implementation of the above approach maxi = 0; ans1 = "";   # Function to check the string def calculate(ans) :           global maxi,ans1;           dp = [ 0 ] * 26;     for i in range(len(ans)) :           # Count the frequency         # of the string         dp[ord(ans[i]) - ord('A')] += 1;       # Check the frequency of the string     for i in range(26) :         if (dp[i] % 2 == 1) :             return;               if (maxi < len(ans)) :           # Store the length         # of the new String         maxi = len(ans);         ans1 = ans;   # Function to find the longest # concatenated string having # every character of even frequency def longestString( arr,  index, string) :         # Checking the string     if (index == len(arr)) :         return;       # Dont Include the string     longestString(arr, index + 1, string);       # Include the string     string += arr[index];       calculate(string);     longestString(arr, index + 1, string);     # Driver code if __name__ == "__main__" :       A = [ "ABAB", "ABF", "CDA", "AD", "CCC" ];         # Call the function     longestString(A, 0, "");       # Print the answer     print(ans1, len(ans1));       # This code is contributed by AnkThon

## C#

 // C# Implementation of the above approach using System;   public class index {     static int max = 0;     static String ans1 = "";       // Function to check the string     static void calculate(String ans)     {           int[] dp = new int[26];         for (int i = 0; i < ans.Length; ++i) {               // Count the frequency             // of the string             dp[(int)ans[i] - (int)'A']++;         }           // Check the frequency of the string         for (int i = 0; i < dp.Length; ++i) {             if (dp[i] % 2 == 1) {                 return;             }         }         if (max < ans.Length) {               // Store the Length             // of the new String             max = ans.Length;             ans1 = ans;         }     }       // Function to find the longest     // concatenated string having     // every character of even frequency     static void longestString(String[] arr, int index, String str)     {           // Checking the string         if (index == arr.Length) {             return;         }           // Dont Include the string         longestString(arr, index + 1, str);           // Include the string         str += arr[index];           calculate(str);         longestString(arr, index + 1, str);     }       // Driver code     public static void Main()     {         String[] A = {"ABAB", "ABF", "CDA", "AD", "CCC"};           // Call the function         longestString(A, 0, "");           // Print the answer         Console.WriteLine(ans1 + " " + ans1.Length);     } }   // This code is contributed by saurabh_jaiswal.

## Javascript



Output

Time Complexity: O(M*N* (2^N)), where N is the number of strings and M is the length of the longest string
Auxiliary Space: O(N)

Another Approach: The above approach can be further optimized by precomputing the frequency of characters for every string and updating the frequency array after concatenation of each string.

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