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Rearrange characters to form palindrome if possible

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  • Difficulty Level : Easy
  • Last Updated : 03 Aug, 2022

Given a string, convert the string to palindrome without any modifications like adding a character, removing a character, replacing a character etc. 

Examples: 

Input : "mdaam"
Output : "madam" or "amdma"

Input : "abb"
Output : "bab"

Input : "geeksforgeeks"
Output : "No Palindrome"
  1. Count occurrences of all characters. 
  2. Count odd occurrences. If this count is greater than 1 or is equal to 1 and length of the string is even then obviously palindrome cannot be formed from the given string. 
  3. Initialize two empty strings firstHalf and secondHalf. 
  4. Traverse the map. For every character with count as count, attach count/2 characters to end of firstHalf and beginning of secondHalf. 
  5. Finally return the result by appending firstHalf and secondHalf

Implementation:

C++




// C++ program to rearrange a string to
// make palindrome.
#include <bits/stdc++.h>
using namespace std;
 
string getPalindrome(string str)
{
 
    // Store counts of characters
    unordered_map<char, int> hmap;
    for (int i = 0; i < str.length(); i++)
        hmap[str[i]]++;
 
    /* find the number of odd elements.
       Takes O(n) */
    int oddCount = 0;
    char oddChar;
    for (auto x : hmap) {
        if (x.second % 2 != 0) {
            oddCount++;
            oddChar = x.first;
        }
    }
 
    /* odd_cnt = 1 only if the length of
       str is odd */
    if (oddCount > 1
        || oddCount == 1 && str.length() % 2 == 0)
        return "NO PALINDROME";
 
    /* Generate first half of palindrome */
    string firstHalf = "", secondHalf = "";
    for (auto x : hmap) {
 
        // Build a string of floor(count/2)
        // occurrences of current character
        string s(x.second / 2, x.first);
 
        // Attach the built string to end of
        // and begin of second half
        firstHalf = firstHalf + s;
        secondHalf = s + secondHalf;
    }
 
    // Insert odd character if there
    // is any
    return (oddCount == 1)
               ? (firstHalf + oddChar + secondHalf)
               : (firstHalf + secondHalf);
}
 
int main()
{
    string s = "mdaam";
    cout << getPalindrome(s);
    return 0;
}

Java




// Java program to rearrange a string to
// make palindrome
import java.util.HashMap;
import java.util.Map.Entry;
 
class GFG{
 
public static String getPalindrome(String str)
{
     
    // Store counts of characters
    HashMap<Character,
            Integer> counting = new HashMap<>();
    for(char ch : str.toCharArray())
    {
        if (counting.containsKey(ch))
        {
            counting.put(ch, counting.get(ch) + 1);
        }
        else
        {
            counting.put(ch, 1);
        }
    }
     
    /* Find the number of odd elements.
    Takes O(n) */
    int oddCount = 0;
    char oddChar = 0;
     
    for(Entry<Character,
              Integer> itr : counting.entrySet())
    {
        if (itr.getValue() % 2 != 0)
        {
            oddCount++;
            oddChar = itr.getKey();
        }
    }
     
    /* odd_cnt = 1 only if the length of
    str is odd */
    if (oddCount > 1 || oddCount == 1 &&
        str.length() % 2 == 0)
    {
        return "NO PALINDROME";
    }
     
    /* Generate first half of palindrome */
    String firstHalf = "", lastHalf = "";
    for(Entry<Character, Integer> itr : counting.entrySet())
    {
         
        // Build a string of floor(count/2)
        // occurrences of current character
        String ss = "";
        for(int i = 0; i < itr.getValue() / 2; i++)
        {
            ss += itr.getKey();
        }
         
        // Attach the built string to end of
        // and begin of second half
        firstHalf = firstHalf + ss;
        lastHalf = ss + lastHalf;
    }
     
    // Insert odd character if there
    // is any
    return (oddCount == 1) ?
           (firstHalf + oddChar + lastHalf) :
           (firstHalf + lastHalf);
}
 
// Driver code
public static void main(String[] args)
{
    String str = "mdaam";
    System.out.println(getPalindrome(str));
}
}
 
// This code is contributed by Satyam Singh

Python3




# Python3 program to rearrange a string to
# make palindrome.
from collections import defaultdict
 
 
def getPalindrome(st):
 
    # Store counts of characters
    hmap = defaultdict(int)
    for i in range(len(st)):
        hmap[st[i]] += 1
 
    # Find the number of odd elements.
    # Takes O(n)
    oddCount = 0
 
    for x in hmap:
        if (hmap[x] % 2 != 0):
            oddCount += 1
            oddChar = x
 
    # odd_cnt = 1 only if the length of
    # str is odd
    if (oddCount > 1 or oddCount == 1 and
            len(st) % 2 == 0):
        return "NO PALINDROME"
 
    # Generate first half of palindrome
    firstHalf = ""
    secondHalf = ""
 
    for x in sorted(hmap.keys()):
 
        # Build a string of floor(count/2)
        # occurrences of current character
        s = (hmap[x] // 2) * x
 
        # Attach the built string to end of
        # and begin of second half
        firstHalf = firstHalf + s
        secondHalf = s + secondHalf
 
    # Insert odd character if there
    # is any
    if (oddCount == 1):
        return (firstHalf + oddChar + secondHalf)
    else:
        return (firstHalf + secondHalf)
 
 
# Driver code
if __name__ == "__main__":
 
    s = "mdaam"
 
    print(getPalindrome(s))
 
# This code is contributed by ukasp

Javascript




<script>
 
// JavaScript program to rearrange a string to
// make palindrome.
function getPalindrome(str)
{
 
    // Store counts of characters
    let hmap = new Map();
    for (let i = 0; i < str.length; i++){
        if(hmap.has(str[i])){
            hmap.set(str[i],hmap.get(str[i])+1);
        }
        else{
            hmap.set(str[i],1);
        }
    }
 
    /* find the number of odd elements.
       Takes O(n) */
    let oddCount = 0;
    let oddChar;
    for (let [x,y] of hmap) {
        if (y % 2 != 0) {
            oddCount++;
            oddChar = x;
        }
    }
 
    /* odd_cnt = 1 only if the length of
       str is odd */
    if (oddCount > 1
        || oddCount == 1 && str.length % 2 == 0)
        return "NO PALINDROME";
 
    /* Generate first half of palindrome */
    let firstHalf = "", secondHalf = "";
    for (let [x,y] of hmap) {
 
        // Build a string of floor(count/2)
        // occurrences of current character
        let s = "";
        for(let i = 0; i < Math.floor(y/2); i++){
             s += x;
        }
 
        // Attach the built string to end of
        // and begin of second half
        firstHalf = firstHalf + s;
        secondHalf = s + secondHalf;
    }
 
    // Insert odd character if there
    // is any
    return (oddCount == 1)
               ? (firstHalf + oddChar + secondHalf)
               : (firstHalf + secondHalf);
}
 
// driver program
let s = "mdaam";
document.write(getPalindrome(s));
 
// This code is contributed by shinjanpatra.
</script>

Output

amdma

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