Check if characters of a given string can be rearranged to form a palindrome
Given a string, Check if characters of the given string can be rearranged to form a palindrome.
For example characters of “geeksogeeks” can be rearranged to form a palindrome “geeksoskeeg”, but characters of “geeksforgeeks” cannot be rearranged to form a palindrome.
A set of characters can form a palindrome if at most one character occurs odd number of times and all characters occur even number of times.
A simple solution is to run two loops, the outer loop picks all characters one by one, the inner loop counts the number of occurrences of the picked character. We keep track of odd counts. Time complexity of this solution is O(n2).
We can do it in O(n) time using a count array. Following are detailed steps.
- Create a count array of alphabet size which is typically 256. Initialize all values of count array as 0.
- Traverse the given string and increment count of every character.
- Traverse the count array and if the count array has more than one odd values, return false. Otherwise, return true.
Below is the implementation of the above approach.
C++
// C++ implementation to check if// characters of a given string can// be rearranged to form a palindrome#include <bits/stdc++.h>using namespace std;#define NO_OF_CHARS 256/* function to check whether characters of a string can form a palindrome */bool canFormPalindrome(string str){ // Create a count array and initialize all // values as 0 int count[NO_OF_CHARS] = { 0 }; // For each character in input strings, // increment count in the corresponding // count array for (int i = 0; str[i]; i++) count[str[i]]++; // Count odd occurring characters int odd = 0; for (int i = 0; i < NO_OF_CHARS; i++) { if (count[i] & 1) odd++; if (odd > 1) return false; } // Return true if odd count is 0 or 1, return true;}/* Driver code*/int main(){ canFormPalindrome("geeksforgeeks") ? cout << "Yes\n" : cout << "No\n"; canFormPalindrome("geeksogeeks") ? cout << "Yes\n" : cout << "No\n"; return 0;} |
Java
// Java implementation to check if// characters of a given string can// be rearranged to form a palindromeimport java.io.*;import java.math.*;import java.util.*;class GFG { static int NO_OF_CHARS = 256; /* function to check whether characters of a string can form a palindrome */ static boolean canFormPalindrome(String str) { // Create a count array and initialize all // values as 0 int count[] = new int[NO_OF_CHARS]; Arrays.fill(count, 0); // For each character in input strings, // increment count in the corresponding // count array for (int i = 0; i < str.length(); i++) count[(int)(str.charAt(i))]++; // Count odd occurring characters int odd = 0; for (int i = 0; i < NO_OF_CHARS; i++) { if ((count[i] & 1) == 1) odd++; if (odd > 1) return false; } // Return true if odd count is 0 or 1, return true; } // Driver code public static void main(String args[]) { if (canFormPalindrome("geeksforgeeks")) System.out.println("Yes"); else System.out.println("No"); if (canFormPalindrome("geeksogeeks")) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Nikita Tiwari. |
Python
# Python3 implementation to check if# characters of a given string can# be rearranged to form a palindromeNO_OF_CHARS = 256# function to check whether characters# of a string can form a palindromedef canFormPalindrome(st): # Create a count array and initialize # all values as 0 count = [0] * (NO_OF_CHARS) # For each character in input strings, # increment count in the corresponding # count array for i in range(0, len(st)): count[ord(st[i])] = count[ord(st[i])] + 1 # Count odd occurring characters odd = 0 for i in range(0, NO_OF_CHARS): if (count[i] & 1): odd = odd + 1 if (odd > 1): return False # Return true if odd count is 0 or 1, return True# Driver codeif(canFormPalindrome("geeksforgeeks")): print("Yes")else: print("No")if(canFormPalindrome("geeksogeeks")): print("Yes")else: print("No")# This code is contributed by Nikita Tiwari. |
C#
// C# implementation to check if// characters of a given string can// be rearranged to form a palindromeusing System;class GFG { static int NO_OF_CHARS = 256; /* function to check whether characters of a string can form a palindrome */ static bool canFormPalindrome(string str) { // Create a count array and initialize all // values as 0 int[] count = new int[NO_OF_CHARS]; Array.Fill(count, 0); // For each character in input strings, // increment count in the corresponding // count array for (int i = 0; i < str.Length; i++) count[(int)(str[i])]++; // Count odd occurring characters int odd = 0; for (int i = 0; i < NO_OF_CHARS; i++) { if ((count[i] & 1) == 1) odd++; if (odd > 1) return false; } // Return true if odd count is 0 or 1, return true; } // Driver code public static void Main() { if (canFormPalindrome("geeksforgeeks")) Console.WriteLine("Yes"); else Console.WriteLine("No"); if (canFormPalindrome("geeksogeeks")) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} |
Output
No Yes
This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Another approach:
We can do it in O(n) time using a list. Following are detailed steps.
- Create a character list.
- Traverse the given string.
- For every character in the string, remove the character if the list already contains else add to the list.
- If the string length is even the list is expected to be empty.
- Or if the string length is odd the list size is expected to be 1
- On the above two conditions (3) or (4) return true else return false.
C++
#include <bits/stdc++.h>using namespace std;/** function to check whether characters ofa string can form a palindrome*/bool canFormPalindrome(string str){ // Create a list vector<char> list; // For each character in input strings, // remove character if list contains // else add character to list for (int i = 0; i < str.length(); i++) { auto pos = find(list.begin(), list.end(), str[i]); if (pos != list.end()) { auto posi = find(list.begin(), list.end(), str[i]); list.erase(posi); } else list.push_back(str[i]); } // if character length is even list is // expected to be empty or if character // length is odd list size is expected to be 1 // if string length is even if (str.length() % 2 == 0 && list.empty() || (str.length() % 2 == 1 && list.size() == 1)) return true; // if string length is odd else return false;}// Driver codeint main(){ if (canFormPalindrome("geeksforgeeks")) cout << ("Yes") << endl; else cout << ("No") << endl; if (canFormPalindrome("geeksogeeks")) cout << ("Yes") << endl; else cout << ("No") << endl;}// This code is contributed by Rajput-Ji |
Java
import java.util.ArrayList;import java.util.List;class GFG { /* * function to check whether * characters of a string can form a palindrome */ static boolean canFormPalindrome(String str) { // Create a list List<Character> list = new ArrayList<Character>(); // For each character in input strings, // remove character if list contains // else add character to list for (int i = 0; i < str.length(); i++) { if (list.contains(str.charAt(i))) list.remove((Character)str.charAt(i)); else list.add(str.charAt(i)); } // if character length is even // list is expected to be empty or // if character length is odd list size // is expected to be 1 // if string length is even if (str.length() % 2 == 0 && list.isEmpty() || (str.length() % 2 == 1 && list.size() == 1)) return true; // if string length is odd else return false; } // Driver code public static void main(String args[]) { if (canFormPalindrome("geeksforgeeks")) System.out.println("Yes"); else System.out.println("No"); if (canFormPalindrome("geeksogeeks")) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Sugunakumar P |
Python
''' * function to check whether characters of a strring can form a palindrome '''def canFormPalindrome(strr): # Create a listt listt = [] # For each character in input strings, # remove character if listt contains # else add character to listt for i in range(len(strr)): if (strr[i] in listt): listt.remove(strr[i]) else: listt.append(strr[i]) # if character length is even # list is expected to be empty # or if character length is odd # listt size is expected to be 1 if (len(strr) % 2 == 0 and len(listt) == 0 or (len(strr) % 2 == 1 and len(listt) == 1)): return True else: return False# Driver codeif (canFormPalindrome("geeksforgeeks")): print("Yes")else: print("No")if (canFormPalindrome("geeksogeeks")): print("Yes")else: print("No")# This code is contributed by SHUBHAMSINGH10 |
C#
// C# Implementation of the above approachusing System;using System.Collections.Generic;class GFG { /* * function to check whether characters of a string can form a palindrome */ static Boolean canFormPalindrome(String str) { // Create a list List<char> list = new List<char>(); // For each character in input strings, // remove character if list contains // else add character to list for (int i = 0; i < str.Length; i++) { if (list.Contains(str[i])) list.Remove((char)str[i]); else list.Add(str[i]); } // if character length is even // list is expected to be empty // or if character length is odd // list size is expected to be 1 // if string length is even if (str.Length % 2 == 0 && list.Count == 0 || (str.Length % 2 == 1 && list.Count == 1)) return true; // if string length is odd else return false; } // Driver Code public static void Main(String[] args) { if (canFormPalindrome("geeksforgeeks")) Console.WriteLine("Yes"); else Console.WriteLine("No"); if (canFormPalindrome("geeksogeeks")) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by Rajput-Ji |
Output
No Yes
Another Approach: (Using Bits)
This problem can be solved in O(n) time where n is the number of characters in the string and O(1) space.
The string to be palindrome all the characters should occur an even number of times if the string is of even length and almost one character can occur an odd number of times if the string length is odd. Track of the count of the characters is not required instead, it is sufficient to keep track if the counts are odd or even.
This can be achieved by using a variable as bit vector.
For every character in the string:
if the bit corresponding to character is not set: //if it is the character’s odd occurrence
set the bit
else if the bit corresponding to character is set: //if it is the character’s even occurrence
toggle the bit
This is similar to performing an XOR operation between bit vector and mask.
Below is the implementation of the above approach:
C++
// C++ Implementation of the above approach# include <bits/stdc++.h>using namespace std;bool canFormPalindrome(string a){ // bitvector to store // the record of which character appear // odd and even number of times int bitvector = 0, mask = 0; for (int i=0; a[i] != '\0'; i++) { int x = a[i] - 'a'; mask = 1 << x; bitvector = bitvector ^ mask; } return (bitvector & (bitvector - 1)) == 0;}// Driver Codeint main(){ if (canFormPalindrome("geeksforgeeks")) cout << ("Yes") << endl; else cout << ("No") << endl; return 0;} |
Python3
# Python implementation of above approach.def canFormPalindrome(s): bitvector = 0 for str in s: bitvector ^= 1 << ord(str) return bitvector == 0 or bitvector & (bitvector - 1) == 0#s = input() if canFormPalindrome("geeksforgeeks"): print('Yes')else: print('No') # This code is contributed by sahilmahale0 |
Output
No Yes
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