Longest substring whose characters can be rearranged to form a Palindrome

Given a string S of length N which only contains lowercase alphabets. Find the length of the longest substring of S such that the characters in it can be rearranged to form a palindrome.

Examples:

Input: S = “aabe”
Output: 3
Explanation:
The substring “aab” can be rearranged to form “aba”, which is a palindromic substring.
Since the length of “aab” is 3 so output is 3.
Notice that “a”, “aa”, “b” and “e” can be arranged to form palindromic strings, but they are not longer than “aab”.

Input: S = “adbabd”
Output: 6
Explanation:
The whole string “adbabd” can be rearranged to form a palindromic substring. One possible arrangement is “abddba”.
Therefore, output length of the string i.e., 6.

Naive Approach: The idea is to generate all possible substring and keep count of each character in it. Initialize answer with the value 0. If the count of each character is even with at most one character with the odd occurrence, the substring can be re-arranged to form a palindromic string. If a substring satisfies this property then update the answer. Print the maximum length after the above steps.

Below is the implementation of the above approach:

C++

// C++ code for the above approach
 
#include <bits/stdc++.h>

using namespace std;
 
// function to check if the string can be
// rearranged to a palindromic string or not

bool ispalindromic(string s)
{
 
    int n = s.size();

    // hashmap to count the frequency of

    // every character in given substring

    unordered_map<char, int> hashmap;
 
    for (auto ch : s) {

        hashmap[ch]++;

    }
 
    int count = 0;
 
    // Count of characters having odd frequency

    for (auto i : hashmap) {

        if (i.second % 2 == 1)

            count++;

    }
 
    // if count is greater than 1

    if (count > 1) {

        return false;

    }
 
    return true;
}
// Function to get the length of longest
// substring whose characters can be
// arranged to form a palindromic string

int longestSubstring(string S, int n)
{

    int ans = 0;
 
    for (int i = 0; i < S.size(); i++) {

        string curstr = "";

        for (int j = i; j < S.size(); j++) {

            // Storing the substring

            curstr += S[j];

              // Checking if it is possible to

            // make it a palindrome

            if (ispalindromic(curstr)

                == true) 

            {

                  // Storing the maximum answer

                ans = max(ans, j - i + 1);

            }

        }

    }
 
    return ans;
}
 
// Driver code

int main()
{
 
    // Given String

    string s = "adbabd";
 
    // Length of given string

    int n = s.size();
 
    // Function call

    cout << (longestSubstring(s, n));
}
 
// This code is contributed by Arpit Jain

Java

// Java code for the above approach

import java.io.*;

import java.util.*;
 
class GFG {
 
  // function to check if the string can be

  // rearranged to a palindromic string or not

  static boolean ispalindromic(String s)

  {
 
    int n = s.length();

    // hashmap to count the frequency of

    // every character in given substring

    HashMap<Character,Integer>hashmap = new HashMap<Character,Integer>();
 
    for (int ch = 0; ch < n; ch++) {

      if(hashmap.containsKey(s.charAt(ch))){

        hashmap.put(s.charAt(ch), hashmap.get(s.charAt(ch))+1);

      }

      else hashmap.put(s.charAt(ch),1);

    }
 
    int count = 0;
 
    // Count of characters having odd frequency

    for(Character i : hashmap.keySet()){

      if (hashmap.get(i) % 2 == 1)

        count++;

    }
 
    // if count is greater than 1

    if (count > 1) {

      return false;

    }
 
    return true;

  }

  // Function to get the length of longest

  // substring whose characters can be

  // arranged to form a palindromic string

  static int longestSubstring(String S, int n)

  {

    int ans = 0;
 
    for (int i = 0; i < S.length(); i++) {

      String curstr = "";

      for (int j = i; j < S.length(); j++) {
 
        // Storing the substring

        curstr += S.charAt(j);
 
        // Checking if it is possible to

        // make it a palindrome

        if (ispalindromic(curstr)

            == true)

        {
 
          // Storing the maximum answer

          ans = Math.max(ans, j - i + 1);

        }

      }

    }
 
    return ans;

  }
 
  // Driver code

  public static void main (String[] args)

  {
 
    // Given String

    String s = "adbabd";
 
    // Length of given string

    int n = s.length();
 
    // Function call

    System.out.println(longestSubstring(s, n));

  }
}
 
// This code is contributed by Aman Kumar.

Python3

# Python code for the above approach
 
# function to check if the string can be
# rearranged to a palindromic string or not

def ispalindromic(s):

    n = len(s)

    # hashmap to count the frequency of

    # every character in given substring

    hashmap={}

    for ch in s:

        if(ch in hashmap):

            hashmap[ch] = hashmap[ch] + 1

        else:

            hashmap[ch] = 1

    count = 0

    # Count of characters having odd frequency

    for key in hashmap:

        if(hashmap[key]%2 == 1):

            count += 1

    # if count is greater than 1

    if(count > 1):

        return False

    return True

# Function to get the length of longest
# substring whose characters can be
#  arranged to form a palindromic string

def longestSubstring(S, n):

    ans = 0

    for i in range(len(S)):

        curstr = ""

        for j in range(i,len(S)):

            # Storing the substring

            curstr += S[j]

            # Checking if it is possible to

            # make it a palindrome

            if(ispalindromic(curstr) == True):

                # Storing the maximum answer

                ans = max(ans, j - i + 1)

    return ans

# Driver code
 
# Given String

s = "adbabd"
 
# Length of given string

n = len(s)
 
# Function call

print(longestSubstring(s,n))
 
# This code is contributed by Pushpesh Raj.

// C# code for the above approach

using System;

using System.Collections.Generic;
 
class GFG {
 
  // function to check if the string can be

  // rearranged to a palindromic string or not

  static bool ispalindromic(string s)

  {
 
    int n = s.Length;
 
    // hashmap to count the frequency of

    // every character in given substring

    Dictionary<char,int> hashmap = new Dictionary<char,int>();
 
    for (int ch = 0; ch < n; ch++) {

      if(hashmap.ContainsKey(s[ch])){

        hashmap[s[ch]] = hashmap[s[ch]]+1;

      }

      else hashmap.Add(s[ch],1);

    }
 
    int count = 0;
 
    // Count of characters having odd frequency

    foreach(KeyValuePair<char,int> i in hashmap){

      if (i.Value % 2 == 1)

        count++;

    }
 
    // if count is greater than 1

    if (count > 1) {

      return false;

    }
 
    return true;

  }
 
  // Function to get the length of longest

  // substring whose characters can be

  // arranged to form a palindromic string

  static int longestSubstring(string S, int n)

  {

    int ans = 0;
 
    for (int i = 0; i < S.Length; i++) {

      string curstr = "";

      for (int j = i; j < S.Length; j++) {
 
        // Storing the substring

        curstr += S[j];
 
        // Checking if it is possible to

        // make it a palindrome

        if (ispalindromic(curstr)

            == true)

        {
 
          // Storing the maximum answer

          ans = Math.Max(ans, j - i + 1);

        }

      }

    }
 
    return ans;

  }
 
  // Driver code

  public static void Main()

  {
 
    // Given String

    string s = "adbabd";
 
    // Length of given string

    int n = s.Length;
 
    // Function call

    Console.WriteLine(longestSubstring(s, n));

  }
}
 
// This code is contributed by Utkarsh

Javascript

// Javascript code for the above approach
 
// function to check if the string can be
// rearranged to a palindromic string or not

function ispalindromic(s) {

  let n = s.length;

  // hashmap to count the frequency of

  // every character in given substring

  let hashmap = {};
 
  for (let i = 0; i < n; i++) {

    if (!hashmap[s[i]]) {

      hashmap[s[i]] = 1;

    } else {

      hashmap[s[i]]++;

    }

  }
 
  let count = 0;
 
  // Count of characters having odd frequency

  for (let key in hashmap) {

    if (hashmap[key] % 2 === 1) count++;

  }
 
  // if count is greater than 1

  if (count > 1) {

    return false;

  }
 
  return true;
}
 
// Function to get the length of longest
// substring whose characters can be
// arranged to form a palindromic string

function longestSubstring(S) {

  let ans = 0;
 
  for (let i = 0; i < S.length; i++) {

    let curstr = "";

    for (let j = i; j < S.length; j++) {

      // Storing the substring

      curstr += S[j];
 
      // Checking if it is possible to

      // make it a palindrome

      if (ispalindromic(curstr) === true) {

        // Storing the maximum answer

        ans = Math.max(ans, j - i + 1);

      }

    }

  }
 
  return ans;
}
 
// Given String

let s = "adbabd";
 
// Length of given string
let n = s.length;
 
// Function call
console.log(longestSubstring(s));
 
// This code is contributed by lokeshpotta20.

Output

Time Complexity: O(N³ * 26)
Auxiliary Space: O(N² * 26)

Efficient Approach: The idea is to observe that the string is a palindrome if at most one character occurs an odd number of times. So there is no need to keep the total count of each character. Just knowing that it is occurring even or an odd number of times is enough. To do this, use bit masking since the count of lowercase alphabets is only 26.

Define a bitmask variable mask which tracks if the occurrence of each character is even or odd.
Create a dictionary index that keeps track of the index of each bitmask.
Traverse the given string S. First, convert the characters from ‘a’ – ‘z’ to 0 – 25 and store this value in a variable temp. For each occurrence of the character, take Bitwise XOR of 2^temp with the mask.
If the character occurs even number of times, its bit in the mask will be off else it will be on. If the mask is currently not in the index, simply assign present index i to bitmask mask in the index.
If the mask is present in the index it means that from the index[mask] to i, the occurrence of all characters is even which is suitable for a palindromic substring. Therefore, update the answer if the length of this segment from the index[mask] to i is greater than the answer.
To check for the substring with one character occurring an odd number of times, iterate a variable j over [0, 25]. Store Bitwise XOR of x with 2^j in mask2.
If mask2 is present in the index, this means that this character is occurring an odd number of times and all characters occur even a number of times in the segment index[mask2] to i, which is also a suitable condition for a palindromic string. Therefore, update our answer with the length of this substring if it is greater than the answer.
Print the maximum length of substring after the above steps.

Below is the implementation of the above approach:

C++

// C++ program for the above approach 
#include<bits/stdc++.h>

using namespace std;

// Function to get the length of longest 
// substring whose characters can be 
// arranged to form a palindromic string 

int longestSubstring(string s, int n) 
{ 

    // To keep track of the last 

    // index of each xor 

    map<int, int> index;

    // Initialize answer with 0 

    int answer = 0; 
 
    int mask = 0; 

    index[mask] = -1; 
 
    // Now iterate through each character 

    // of the string 

    for(int i = 0; i < n; i++) 

    { 

        // Convert the character from 

        // [a, z] to [0, 25] 

        int temp = (int)s[i] - 97; 
 
        // Turn the temp-th bit on if 

        // character occurs odd number 

        // of times and turn off the temp-th 

        // bit off if the character occurs 

        // ever number of times 

        mask ^= (1 << temp); 
 
        // If a mask is present in the index 

        // Therefore a palindrome is 

        // found from index[mask] to i 

        if (index[mask]) 

        { 

            answer = max(answer, 

                         i - index[mask]); 

        } 
 
        // If x is not found then add its 

        // position in the index dict. 

        else

            index[mask] = i; 
 
        // Check for the palindrome of 

        // odd length 

        for(int j = 0; j < 26; j++) 

        { 

            // We cancel the occurrence 

            // of a character if it occurs 

            // odd number times 

            int mask2 = mask ^ (1 << j); 

            if (index[mask2]) 

            { 

                answer =max(answer, 

                            i - index[mask2]); 

            } 

        } 

    } 

    return answer; 
} 

// Driver code 

int main () 
{ 

    // Given String 

    string s = "adbabd"; 

    // Length of given string 

    int n = s.size(); 

    // Function call 

    cout << (longestSubstring(s, n)); 
}
 
// This code is contributed by Stream_Cipher

Java

// Java program for the above approach 

import java.util.*;
 
class GFG{

// Function to get the length of longest 
// substring whose characters can be 
// arranged to form a palindromic string 

static int longestSubstring(String s, int n)
{

    // To keep track of the last 

    // index of each xor 

    Map<Integer, Integer> index = new HashMap<>();
 
    // Initialize answer with 0 

    int answer = 0;
 
    int mask = 0;

    index.put(mask, -1);
 
    // Now iterate through each character 

    // of the string 

    for(int i = 0; i < n; i++)

    {
 
        // Convert the character from 

        // [a, z] to [0, 25] 

        int temp = (int)s.charAt(i) - 97;
 
        // Turn the temp-th bit on if 

        // character occurs odd number 

        // of times and turn off the temp-th 

        // bit off if the character occurs 

        // ever number of times 

        mask ^= (1 << temp);
 
        // If a mask is present in the index 

        // Therefore a palindrome is 

        // found from index[mask] to i 

        if (index.containsKey(mask))

        { 

            answer = Math.max(answer, 

                i - index.get(mask)); 

        }
 
        // If x is not found then add its 

        // position in the index dict. 

        else

            index.put(mask,i); 
 
        // Check for the palindrome of 

        // odd length 

        for (int j = 0;j < 26; j++)

        {

            // We cancel the occurrence 

            // of a character if it occurs 

            // odd number times 

            int mask2 = mask ^ (1 << j);

            if (index.containsKey(mask2))

            { 

                answer = Math.max(answer, 

                    i - index.get(mask2)); 

            }

        }

    }

    return answer;
}

// Driver code

public static void main (String[] args)
{

    // Given String 

    String s = "adbabd";

    // Length of given string 

    int n = s.length();

    // Function call 

    System.out.print(longestSubstring(s, n));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to get the length of longest
# substring whose characters can be
# arranged to form a palindromic string

def longestSubstring(s: str, n: int):
 
    # To keep track of the last

    # index of each xor

    index = dict()
 
    # Initialize answer with 0

    answer = 0
 
    mask = 0

    index[mask] = -1
 
    # Now iterate through each character

    # of the string

    for i in range(n):
 
        # Convert the character from

        # [a, z] to [0, 25]

        temp = ord(s[i]) - 97
 
        # Turn the temp-th bit on if

        # character occurs odd number

        # of times and turn off the temp-th

        # bit off if the character occurs

        # ever number of times

        mask ^= (1 << temp)
 
        # If a mask is present in the index

        # Therefore a palindrome is

        # found from index[mask] to i

        if mask in index.keys():

            answer = max(answer,

                         i - index[mask])
 
        # If x is not found then add its

        # position in the index dict.

        else:

            index[mask] = i
 
        # Check for the palindrome of

        # odd length

        for j in range(26):
 
            # We cancel the occurrence

            # of a character if it occurs

            # odd number times

            mask2 = mask ^ (1 << j)

            if mask2 in index.keys():
 
                answer = max(answer,

                             i - index[mask2])
 
    return answer
 
# Driver Code
 
# Given String

s = "adbabd"
 
# Length of given string

n = len(s)
 
# Function call

print(longestSubstring(s, n))

// C# program for the above approach 

using System.Collections.Generic; 

using System; 
 
class GFG{ 

// Function to get the length of longest 
// substring whose characters can be 
// arranged to form a palindromic string 

static int longestSubstring(string s, int n) 
{ 

    // To keep track of the last 

    // index of each xor 

    Dictionary<int, 

               int> index = new Dictionary<int,

                                           int>();

    // Initialize answer with 0 

    int answer = 0; 
 
    int mask = 0; 

    index[mask] = -1; 
 
    // Now iterate through each character 

    // of the string 

    for(int i = 0; i < n; i++) 

    { 
 
        // Convert the character from 

        // [a, z] to [0, 25] 

        int temp = (int)s[i] - 97; 
 
        // Turn the temp-th bit on if 

        // character occurs odd number 

        // of times and turn off the temp-th 

        // bit off if the character occurs 

        // ever number of times 

        mask ^= (1 << temp); 
 
        // If a mask is present in the index 

        // Therefore a palindrome is 

        // found from index[mask] to i 

        if (index.ContainsKey(mask) == true) 

        { 

            answer = Math.Max(answer, 

                              i - index[mask]); 

        } 
 
        // If x is not found then add its 

        // position in the index dict. 

        else

            index[mask] = i; 
 
        // Check for the palindrome of 

        // odd length 

        for(int j = 0; j < 26; j++) 

        { 

            // We cancel the occurrence 

            // of a character if it occurs 

            // odd number times 

            int mask2 = mask ^ (1 << j); 

            if (index.ContainsKey(mask2) == true) 

            { 

                answer = Math.Max(answer, 

                                  i - index[mask2]); 

            } 

        } 

    } 

    return answer; 
} 

// Driver code 

public static void Main () 
{ 

    // Given String 

    string s = "adbabd"; 

    // Length of given string 

    int n = s.Length; 

    // Function call 

    Console.WriteLine(longestSubstring(s, n)); 
} 
} 
 
// This code is contributed by Stream_Cipher

Javascript

<script>
 
// Javascript program for the above approach 

// Function to get the length of longest 
// substring whose characters can be 
// arranged to form a palindromic string 

function longestSubstring(s, n) 
{ 

    // To keep track of the last 

    // index of each xor 

    var index = new Map();

    // Initialize answer with 0 

    var answer = 0; 
 
    var mask = 0; 

    index[mask] = -1; 
 
    // Now iterate through each character 

    // of the string 

    for(var i = 0; i < n; i++) 

    { 

        // Convert the character from 

        // [a, z] to [0, 25] 

        var temp = s[i].charCodeAt(0) - 97; 
 
        // Turn the temp-th bit on if 

        // character occurs odd number 

        // of times and turn off the temp-th 

        // bit off if the character occurs 

        // ever number of times 

        mask ^= (1 << temp); 
 
        // If a mask is present in the index 

        // Therefore a palindrome is 

        // found from index[mask] to i 

        if (index[mask]) 

        { 

            answer = Math.max(answer, 

                         i - index[mask]); 

        } 
 
        // If x is not found then add its 

        // position in the index dict. 

        else

            index[mask] = i; 
 
        // Check for the palindrome of 

        // odd length 

        for(var j = 0; j < 26; j++) 

        { 

            // We cancel the occurrence 

            // of a character if it occurs 

            // odd number times 

            var mask2 = mask ^ (1 << j); 

            if (index[mask2]) 

            { 

                answer = Math.max(answer, 

                            i - index[mask2]); 

            } 

        } 

    } 

    return answer; 
} 

// Driver code 
// Given String 

var s = "adbabd"; 
 
// Length of given string 

var n = s.length; 
 
// Function call 
document.write(longestSubstring(s, n)); 
 
// This code is contributed by rrrtnx.
</script>