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 notbool 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 stringint 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 codeint 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 approachimport 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 notdef 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 stringdef 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 Strings = "adbabd"# Length of given stringn = len(s)# Function callprint(longestSubstring(s,n))# This code is contributed by Pushpesh Raj. |
C#
// C# code for the above approachusing 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 notfunction 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 stringfunction 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 Stringlet s = "adbabd";// Length of given stringlet n = s.length;// Function callconsole.log(longestSubstring(s));// This code is contributed by lokeshpotta20. |
Output
6
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 2temp 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 2j 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 codepublic 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 stringdef 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 Strings = "adbabd"# Length of given stringn = len(s)# Function callprint(longestSubstring(s, n)) |
C#
// 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> |
Output
6
Time Complexity: O(N * 26)
Auxiliary Space: O(N * 26)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
