Given N(very large), task if to print the largest palindromic number obtained by permuting the digits of N. If it is not possible to make a palindromic number, then print an appropriate message.

**Examples :**

Input : 313551 Output : 531135 Explanations : 531135 is the largest number which is a palindrome, 135531, 315513 and other numbers can also be formed but we need the highest of all of the palindromes. Input : 331 Output : 313 Input : 3444 Output : Pallindrome cannot be formed

**Naive Approach : ** The **naive** approach will be to try all the permutations possible, and print the largest of such combinations which is a palindrome.

**Efficient Approach :** An efficient approach will be to use **Greedy algorithm.** Since the number is large, store the number in a string. Store the count of occurrences of every digit in the given number in a map. Check if it is possible to form a palindrome or not. If the digits of the given number can be rearranged to form a palindrome, then apply the greedy approach to obtain the number. Check for the occurrence of every digit (9 to 0), and place every available digit at front and back. **Initially, the front pointer will be at index 0, as the largest digit will be placed at first to make the number a large one.** With every step, move the front pointer 1 position ahead. If the digit occurs an odd number of times, then place **one digit in the middle and rest of the even number of digits at front and back**. Keep repeating the process **(map[digit]/2)** number of times for a single digit. After placing a particular digit which occurs an even number of times at the front and back, move the front pointer one step ahead. The placing is done till map[digit] is 0. The char array will have the largest palindromic number possible after completion of the placing of digits greedily.

In the worst case, the time complexity will be **O(10 * (length of string/2))**, in case the number consists of a same digit at every position.

Below is the implementation of the above idea :

// CPP program to print the largest palindromic // number by permuting digits of a number #include <bits/stdc++.h> using namespace std; // function to check if a number can be // permuted to form a palindrome number bool possibility(unordered_map<int, int> m, int length, string s) { // counts the occurrence of number which is odd int countodd = 0; for (int i = 0; i < length; i++) { // if occurrence is odd if (m[s[i] - '0'] & 1) countodd++; // if number exceeds 1 if (countodd > 1) return false; } return true; } // function to print the largest palindromic number // by permuting digits of a number void largestPalindrome(string s) { // string length int l = s.length(); // map that marks the occurrence of a number unordered_map<int, int> m; for (int i = 0; i < l; i++) m[s[i] - '0']++; // check the possibility of a palindromic number if (possibility(m, l, s) == false) { cout << "Palindrome cannot be formed"; return; } // string array that stores the largest // permuted palindromic number char largest[l]; // pointer of front int front = 0; // greedily start from 9 to 0 and place the // greater number in front and odd in the // middle for (int i = 9; i >= 0; i--) { // if the occurrence of number is odd if (m[i] & 1) { // place one odd occurring number // in the middle largest[l / 2] = char(i + 48); // decrease the count m[i]--; // place the rest of numbers greedily while (m[i] > 0) { largest[front] = char(i + 48); largest[l - front - 1] = char(i + 48); m[i] -= 2; front++; } } else { // if all numbers occur even times, // then place greedily while (m[i] > 0) { // place greedily at front largest[front] = char(i + 48); largest[l - front - 1] = char(i + 48); // 2 numbers are placed, so decrease the count m[i] -= 2; // increase placing position front++; } } } // print the largest string thus formed for (int i = 0; i < l; i++) cout << largest[i]; } // Driver Code int main() { string s = "313551"; largestPalindrome(s); return 0; }

**Output:**

531135

**Time Complexity : **O(n), where n is the length of string.

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