Rearrange characters to form palindrome if possible

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

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





// CPP 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++)
  /* find the number of odd elements.
     Takes O(n) */
  int oddCount = 0;
  char oddChar;
  for (auto x : hmap) {
    if (x.second % 2 != 0) {
      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 halh 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;




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