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:
Input: N = 5, str = [“ABAB”, “ABF”, “CDA”, “AD”, “CCC”]
Output: ABABCDAADCCC 12
Explanation: The string formed by concatenation is ABABCDAADCCC. Each letter in the string occurs even number of timesInput: 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
- Include the string and check if the frequency of characters is even for the concatenated string
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;int maxi = 0;string ans1 = "";// Function to check the stringvoid 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 frequencyvoid longestString(vector<string> 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 codeint main(){ vector<string> 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 approachimport 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<String> 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<String> 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 approachmaxi = 0;ans1 = "";# Function to check the stringdef 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 frequencydef 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 codeif __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 approachusing 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
<script>// Javascript implementation of the above approachlet maxi = 0;let ans1 = "";// Function to check the stringfunction calculate(ans) { let dp = new Array(26).fill(0); for (let i = 0; i < ans.length; ++i) { // Count the frequency // of the string dp[ans[i].charCodeAt(0) - "A".charCodeAt(0)]++; } // Check the frequency of the string for (let 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 frequencyfunction longestString(arr, index, 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 codelet A = ["ABAB", "ABF", "CDA", "AD", "CCC"];// Call the functionlongestString(A, 0, "");// Print the answerdocument.write(ans1 + " " + ans1.length);// This code is contributed by gfgking</script> |
Output
ABABCDAADCCC 12
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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