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The maximum length of string formed by concatenation having an even frequency of each character
Concatenation is an operator which is used to join one or more than one string to originate a new string which would be a combination of strings that were used to generate it with the help of concatenation. In the following article, we will take the uppercase letters only in the input string.
Concatenation is an operator which is used to join one or more than one string to originate a new string which would be a combination of strings that were used to generate it with the help of concatenation. In the following article, we will take the uppercase letters only in the input string.
Let’s try to understand this problem with the help of some examples.
Input
"FEFEE", "EFF", "HGG", "AD", "HHH"
Explanation − In this example, we can see that the maximum length of string which can be formed by concatenation having an even frequency of each character is “FEFEEEFFHGGHHH” and the string we get has a length of 14. Thus, ‘14’ is the final output.
Frequency of F− 4
Frequency of E− 4
Frequency of H− 4
Frequency of G− 2
Input
"ABCD", "AVP", "ADDDC", "BCCCA", "HH", "CA"
Output
18
Explanation − In this example, we can see that the maximum length of string which can be formed by concatenation having an even frequency of each character is “ABCDADDDCBCCCAHHCA” and the string we get has a length of 18. Thus, ‘18’ is the final output.
Frequency of A: 4
Frequency of B: 2
Frequency of C: 6
Frequency of D: 4
Frequency of H: 2
Problem Explanation
Let’s try to understand the problem and find its solution. In the following article, we will discuss the approach to find the maximum length of string formed by concatenation having an even frequency of each character of the strings from the given array of strings. The solution will involve recursion and backtracking. For each string in the array, we will have two choices - either to include it into our final string having an even frequency of each character or we would not include that string.
So, the approach involves creating a recursive function and calling it recursively twice for each string - one after including it and another without including it. After including each string, keep checking whether all the characters in the final string have an even frequency, and keep updating the maximum length of a string having an even frequency of each character formed by concatenation.
Recursive Algorithm
Make a recursive function with the base case as when the index would be equal to the size of the array, the function will return. Now, in that function, we will take two cases, in the first case, we would not include the current string and make the recursive call.
While in the second case, we would include it and then check if the present output string matches the requirements by using another function (CheckValid), that is, it is having even frequency of each character or not
If yes, we will store the length of that string if it is greater than the previous size.
Next, we will make another recursive call in which we will include the present string.
This way, we will reach the final output size.
Example
The following are implementations of the above approach in various programming languages −
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
// Function to check whether the string has an even frequency of each character or not
bool CheckValid(const char *str){
// Declare a frequency array that would work like a map to store the count of each alphabet
int mapArray[26] = { 0 };
// Store the frequency of each alphabet, we will take uppercase English alphabets only in our string
for (int i = 0; i < strlen(str); i++) {
mapArray[str[i] - 'A']++;
}
// Check if the frequency of any alphabet is odd, if yes return false
for (int i = 0; i < 26; i++) {
if (mapArray[i] % 2 == 1) {
return false;
}
}
// If we have all our alphabets in even count, we can return true
return true;
}
// Function to find the maximum length of a string having an even frequency of each character formed by concatenation
void FindMaxLength(const char *array[], int i, int size, const char *s, int* maxi){
// Check if we have taken all strings from our array that is check the base case
if (i == size) {
return;
}
// Do not Include the string
FindMaxLength(array, i + 1, size, s, maxi);
// Include the string
char combined[100]; // Choose a reasonable maximum length for your combined string
strcpy(combined, s);
strcat(combined, array[i]);
if (CheckValid(combined)) {
int currentLength = strlen(combined);
if (currentLength > *maxi) {
*maxi = currentLength;
}
}
FindMaxLength(array, i + 1, size, combined, maxi);
}
int main(){
// Size of the string provided by the user
int size = 5;
// String provided by the user
const char* array[5] = { "FEFEE", "EFF", "HGG", "AD", "HHH" };
// Declare a variable to store the maximum length of string formed by concatenation having even frequency of each character
int maxi = 0;
// Call the function to find the maximum length of string formed by concatenation having an even frequency of each character
FindMaxLength(array, 0, size, "", &maxi);
// Print the final output
printf("The maximum length of string formed by concatenation having an even frequency of each character is: %d", maxi);
return 0;
}
Output
The maximum length of string formed by concatenation having an even frequency of each character is: 14
#include <bits/stdc++.h>
using namespace std;
// Function to check whether the string has an even frequency of each character or not
bool CheckValid(string str){
// Declare a frequency array that would work like a map to store the count of each alphabet
int mapArray[26] = { 0 };
// Store the frequency of each alphabet, we will take uppercase English alphabets only in our string
for (int i = 0; i < str.length(); i++) {
mapArray[str[i] - 'A']++;
}
// Check if the frequency of any alphabet is odd, if yes return false
for (int i = 0; i < 26; i++) {
if (mapArray[i] % 2 == 1) {
return false;
}
}
// If we have all our alphabets in even count, we can return true
return true;
}
// Function to find the maximum length of a string having an even frequency of each character formed by concatenation
void FindMaxLength(string array[], int i, int size, string s, int& maxi){
// Check if we have taken all strings from our array that is check the base case
if (i == size) {
return;
}
// Do not Include the string
FindMaxLength(array, i + 1, size, s, maxi);
// Include the string
s += array[i];
// Check if the string collected is giving us a string with all character counts as even, constituted in it
if(CheckValid(s)) {
// Store the maximum of the previous and current length
if(s.length() > maxi) {
maxi = s.length();
}
}
FindMaxLength(array, i + 1, size, s, maxi);
}
int main(){
// Size of the string provided by the user
int size = 5;
// String provided by the user
string array[5] = { "FEFEE", "EFF", "HGG", "AD", "HHH" };
// Declare a variable to store the maximum length of string formed by concatenation having even frequency of each character
int maxi = 0;
// Call the function to find the maximum length of string formed by concatenation having an even frequency of each character
FindMaxLength(array, 0, size, "", maxi);
// Print the final output
cout << "The maximum length of string formed by concatenation having an even frequency of each character is: "<< maxi <<endl;
return 0;
}
Output
The maximum length of string formed by concatenation having an even frequency of each character is: 14
public class Main {
public static boolean CheckValid(String str) {
// Declare a frequency array that would work like a map to store the count of each alphabet
int[] mapArray = new int[26];
// Store the frequency of each alphabet, we will take uppercase English alphabets only in our string
for (int i = 0; i < str.length(); i++) {
mapArray[str.charAt(i) - 'A']++;
}
// Check if the frequency of any alphabet is odd, if yes return false
for (int i = 0; i < 26; i++) {
if (mapArray[i] % 2 == 1) {
return false;
}
}
// If we have all our alphabets in even count, we can return true
return true;
}
//Function to find the maximum length of a string having an even frequency of each character formed by concatenation
public static void FindMaxLength(String[] array, int i, int size, String s, int[] maxi) {
// Check if we have taken all strings from our array that is check the base case
if (i == size) {
return;
}
FindMaxLength(array, i + 1, size, s, maxi);
s += array[i];
if (CheckValid(s)) {
int currentLength = s.length();
// Store the maximum of the previous and current length
if (currentLength > maxi[0]) {
maxi[0] = currentLength;
}
}
FindMaxLength(array, i + 1, size, s, maxi);
}
public static void main(String[] args) {
int size = 5;
String[] array = { "FEFEE", "EFF", "HGG", "AD", "HHH" };
int[] maxi = { 0 };
FindMaxLength(array, 0, size, "", maxi);
System.out.println("The maximum length of a string formed by concatenation having an even frequency of each character is: " + maxi[0]);
}
}
Output
The maximum length of a string formed by concatenation having an even frequency of each character is: 14
# Function to check whether the string has an even frequency of each character or not
def CheckValid(s):
# Declare a frequency array that would work like a map to store the count of each alphabet
mapArray = [0] * 26
# Store the frequency of each alphabet; we will take uppercase English alphabets only in our string
for char in s:
if 'A' <= char <= 'Z':
mapArray[ord(char) - ord('A')] += 1
# Check if the frequency of any alphabet is odd; if yes, return False
for count in mapArray:
if count % 2 == 1:
return False
# If we have all our alphabets in even count, we can return True
return True
# Function to find the maximum length of a string having an even frequency of each character formed by concatenation
def FindMaxLength(array, i, size, s, maxi):
# Check if we have taken all strings from our array, that is, check the base case
if i == size:
return
# Do not include the string
FindMaxLength(array, i + 1, size, s, maxi)
# Include the string
s += array[i]
# Check if the string collected is giving us a string with all character counts as even, constituted in it
if CheckValid(s):
# Store the maximum of the previous and current length
if len(s) > maxi[0]:
maxi[0] = len(s)
FindMaxLength(array, i + 1, size, s, maxi)
if __name__ == "__main__":
size = 5
# String provided by the user
array = ["FEFEE", "EFF", "HGG", "AD", "HHH"]
# Declare a variable to store the maximum length of string formed by concatenation having even frequency of each character
maxi = [0]
# Call the function to find the maximum length of string formed by concatenation having an even frequency of each character
FindMaxLength(array, 0, size, "", maxi)
# Print the final output
print("The maximum length of a string formed by concatenation having an even frequency of each character is:", maxi[0])
Output
The maximum length of a string formed by concatenation having an even frequency of each character is: 14
Complexities for the Above Code
Time complexity − O(n * m * (2^n))); where ‘n’ is the length of the array that is the total number of strings given and ‘m’ is the length of the longest string present in the array. Each string will be executed twice in the recursive call making the time complexity of that recursive function ‘2^ n’ while the CheckValid function will take the total time of the size of the maximum length of the string formed by concatenation and it could vary up to (n * m), that is the case when we will include each string contained in given array in our final answer.
Space complexity − O(1); We have not stored any variable in some data structure in the above code.
Conclusion
In this article, to find the maximum length of string formed by concatenation having even frequency of each character. We understood the approach by using concepts like recursion and backtracking. However, the time complexity of the code given above is huge so the code would work only within some prohibited constraints when the size of string and array will have a limit.
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