Maximum even length sub-string that is permutation of a palindrome

Given string , the task is to find the maximum length of the sub-string of that can be arranged into a Palindrome (i.e at least one of its permutation is a Palindrome). 

Note that the sub-string must be of even length. 

Examples:

Input: str = “124565463” 
Output: 6 “456546” is the valid sub-string 

Input: str = “122313” 
Output: 6

Approach:

Use two variables: start (inclusive) and end (exclusive) to keep track of the starting and ending index of the current sub-string that is being considered in the given string. Also use a dictionary named count which keeps the record of how many times, a character occurs in the current sub-string. 

Now, there are two possible cases for a sub-string:

1. If the length of the sub-string is odd, then it cannot be considered in the final solution.
2. If the length of the sub-string is even, then it can be a possible solution only if each character in that sub-string occurs even number of times which can be done using the dictionary count. We check if each character occurs an even number of times or not. If yes, then we include it as one of the possible solutions. Then we form the next sub-string by including the next character in the string which can be done by simply incrementing the end and check recursively if a sub-string with greater length can be formed which satisfies the given conditions and return the maximum of all possible solutions.

Below is the implementation of the above approach: 

C++

// C++ code to find the maximum length of
// sub-string (of even length) which can be
// arranged into a Palindrome
#include <bits/stdc++.h>
using namespace std;

unordered_map<int, int> countt;

// function that returns true if the given
// sub-string can be arranged into a Palindrome
bool canBePalindrome(unordered_map<int, int> &countt)
{
    for (auto key : countt)
    {
        if (key.second & 1) return false;
    }
    return true;
}

// This function returns the maximum length of
// the sub-string (of even length) which can be
// arranged into a Palindrome
int maxPal(string str,
           unordered_map<int, int> &countt,
           int start, int end)
{

    // If we reach end of the string
    if (end == str.length())
    {

        // if string is of even length
        if ((end - start) % 2 == 0)

            // if string can be arranged into a
            // palindrome
            if (canBePalindrome(countt)) return end - start;

        return 0;
    }
    else
    {

        // Even length sub-string
        if ((end - start) % 2 == 0)
        {

            // Check if current sub-string can be
            // arranged into a palindrome
            if (canBePalindrome(countt))
            {
                countt[str[end]]++;
                return max(end - start, maxPal(str, countt,
                                               start, end + 1));
            }
            else
            {
                countt[str[end]]++;
                return maxPal(str, countt, start, end + 1);
            }
        }

        // Odd length sub-string
        else
        {
            countt[str[end]]++;
            unordered_map<int, int> c(countt.begin(),
                                      countt.end());
            int length = maxPal(str, c, start, end + 1);

            countt[str[end]]--;
            countt[str[start]]--;
            return max(length, maxPal(str, countt,
                                      start + 1, end));
        }
    }
}

// Driver code
int main(int argc, char const *argv[])
{
    string str = "124565463";
    int start = 0, end = 0;

    cout << maxPal(str, countt, start, end) << endl;
    return 0;
}

// This code is contributed by
// sanjeev2552

Java

// Java code to find the maximum length of
// sub-string (of even length) which can be
// arranged into a Palindrome
import java.io.*;
import java.util.*;

class GFG
{

    static HashMap<Integer, Integer> count = new HashMap<>();

    // function that returns true if the given
    // sub-string can be arranged into a Palindrome
    static boolean canBePalindrome(HashMap<Integer, Integer> count)
    {
        for (HashMap.Entry<Integer, Integer> entry : count.entrySet())
            if ((entry.getValue() & 1) == 1)
                return false;
        return true;
    }

    // This function returns the maximum length of
    // the sub-string (of even length) which can be
    // arranged into a Palindrome
    static int maxPal(String str, int start, int end,
                       HashMap<Integer, Integer> count)
    {

        // If we reach end of the string
        if (end == str.length())
        {

            // if string is of even length
            if ((end - start) % 2 == 0)

                // if string can be arranged into a
                // palindrome
                if (canBePalindrome(count))
                    return end - start;

            return 0;
        }
        else
        {

            // Even length sub-string
            if ((end - start) % 2 == 0)
            {

                // Check if current sub-string can be
                // arranged into a palindrome
                if (canBePalindrome(count))
                {
                    count.put((int) str.charAt(end),
                    count.get((int) str.charAt(end)) ==
                    null ? 1 : count.get((int) str.charAt(end)) + 1);
                    return Math.max(end - start,
                            maxPal(str, start, end + 1, count));
                }
                else
                {
                    count.put((int) str.charAt(end),
                    count.get((int) str.charAt(end)) ==
                    null ? 1 : count.get((int) str.charAt(end)) + 1);
                    return maxPal(str, start, end + 1, count);
                }
            }

            // Odd length sub-string
            else
            {
                count.put((int) str.charAt(end),
                count.get((int) str.charAt(end)) ==
                null ? 1 : count.get((int) str.charAt(end)) + 1);
                HashMap<Integer, Integer> c = new HashMap<>(count);

                int length = maxPal(str, start, end + 1, c);

                count.put((int) str.charAt(end),
                count.get((int) str.charAt(end)) ==
                null ? -1 : count.get((int) str.charAt(end)) - 1);

                count.put((int) str.charAt(start),
                count.get((int) str.charAt(start)) ==
                null ? -1 : count.get((int) str.charAt(start)) - 1);

                return Math.max(length, maxPal(str,
                            start + 1, end, count));
            }
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        String str = "124565463";
        int start = 0, end = 0;
        System.out.println(maxPal(str, start, end, count));
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 code to find the maximum length of sub-string
# (of even length) which can be arranged into a Palindrome

from collections import defaultdict

# function that returns true if the given
# sub-string can be arranged into a Palindrome
def canBePalindrome(count):
    for key in count:
        if count[key] % 2 != 0:
            return False
    return True

# This function returns the maximum length of
# the sub-string (of even length) which can be
# arranged into a Palindrome
def maxPal(string, count, start, end):

    # If we reach end of the string
    if end == len(string):

        # if string is of even length
        if (end-start) % 2 == 0:

            # if string can be arranged into a
            # palindrome
            if canBePalindrome(count) == True:
                return end-start

        return 0

    else:

        # Even length sub-string
        if (end-start) % 2 == 0:

            # Check if current sub-string can be
            # arranged into a palindrome
            if canBePalindrome(count) == True:
                count[string[end]] += 1
                return max(end-start, maxPal(string, count, start, end + 1))

            else:
                count[string[end]] += 1
                return maxPal(string, count, start, end + 1)

        # Odd length sub-string
        else:

            count[string[end]] += 1
            length = maxPal(string, count.copy(), start, end + 1)

            count[string[end]] -= 1
            count[string[start]] -= 1
            return max(length, maxPal(string, count, start + 1, end))

# Driver code
string = '124565463'
start, end = 0, 0
count = defaultdict(lambda : 0)

print(maxPal(string, count, start, end))

C#

using System;
using System.Collections.Generic;

public class GFG
{
  static Dictionary<int, int> count = new Dictionary<int, int>();

  // function that returns true if the given
  // sub-string can be arranged into a Palindrome
  static bool CanBePalindrome(Dictionary<int, int> count)
  {
    foreach (KeyValuePair<int, int> entry in count)
      if ((entry.Value & 1) == 1)
        return false;
    return true;
  }

  // This function returns the maximum length of
  // the sub-string (of even length) which can be
  // arranged into a Palindrome
  static int MaxPal(string str, int start, int end,
                    Dictionary<int, int> count)
  {
    // If we reach end of the string
    if (end == str.Length)
    {
      // if string is of even length
      if ((end - start) % 2 == 0)
      {
        // if string can be arranged into a
        // palindrome
        if (CanBePalindrome(count))
          return end - start;

        return 0;
      }
      else
      {
        return 0;
      }
    }
    else
    {
      // Even length sub-string
      if ((end - start) % 2 == 0)
      {
        // Check if current sub-string can be
        // arranged into a palindrome
        if (CanBePalindrome(count))
        {
          count[str[end]] = count.ContainsKey(str[end]) ?
            count[str[end]] + 1 : 1;
          return Math.Max(end - start,
                          MaxPal(str, start, end + 1, count));
        }
        else
        {
          count[str[end]] = count.ContainsKey(str[end]) ?
            count[str[end]] + 1 : 1;
          return MaxPal(str, start, end + 1, count);
        }
      }
      // Odd length sub-string
      else
      {
        count[str[end]] = count.ContainsKey(str[end]) ?
          count[str[end]] + 1 : 1;
        Dictionary<int, int> c = new Dictionary<int, int>(count);

        int length = MaxPal(str, start, end + 1, c);

        count[str[end]] = count.ContainsKey(str[end]) ?
          count[str[end]] - 1 : -1;

        count[str[start]] = count.ContainsKey(str[start]) ?
          count[str[start]] - 1 : -1;

        return Math.Max(length, MaxPal(str,
                                       start + 1, end, count));
      }
    }
  }

  // Driver Code
  public static void Main()
  {
    string str = "124565463";
    int start = 0, end = 0;
    Console.WriteLine(MaxPal(str, start, end, count));
  }
}

Javascript

// Javacript program for the above approach

// function that returns true if the given
// sub-string can be arranged into a Palindrome
function canBePalindrome(count) {
  for (let key in count) {
    if (count[key] % 2 !== 0) {
      return false;
    }
  }
  return true;
}

// This function returns the maximum length of
// the sub-string (of even length) which can be
// arranged into a Palindrome
function maxPal(string, count, start, end)
{

  // If we reach end of the string
  if (end === string.length) {

    // if string is of even length
    if ((end - start) % 2 === 0) {

      // if string can be arranged into a palindrome
      if (canBePalindrome(count) === true) {
        return end - start;
      }
      return 0;
    } else {
      return 0;
    }
  } else {
    // Even length sub-string
    if ((end - start) % 2 === 0) {

      // Check if current sub-string can be
      // arranged into a palindrome
      if (canBePalindrome(count) === true) {
        count[string[end]] = (count[string[end]] || 0) + 1;
        return Math.max(end - start, maxPal(string, {...count}, start, end + 1));
      } else {
        count[string[end]] = (count[string[end]] || 0) + 1;
        return maxPal(string, {...count}, start, end + 1);
      }
    } else {
      count[string[end]] = (count[string[end]] || 0) + 1;
      let length = maxPal(string, {...count}, start, end + 1);
      count[string[end]] -= 1;
      count[string[start]] = (count[string[start]] || 0) - 1;
      return Math.max(length, maxPal(string, {...count}, start + 1, end));
    }
  }
}

// Driver code
let string = '124565463';
let start = 0;
let end = 0;
let count = {};

console.log(maxPal(string, count, start, end));

// This code is contributed by codebraxnzt

Output

6

Time Complexity: O(n * 2^n)

Auxiliary space: O(n)