# Largest palindromic number by permuting digits

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 :

## C++

`// 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; ` `} ` |

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## Python3

# Python3 program to print the largest palindromic

# number by permuting digits of a number

from collections import defaultdict

# Function to check if a number can be

# permuted to form a palindrome number

def possibility(m, length, s):

# counts the occurrence of

# number which is odd

countodd = 0

for i in range(0, length):

# if occurrence is odd

if m[int(s[i])] & 1:

countodd += 1

# if number exceeds 1

if countodd > 1:

return False

return True

# Function to print the largest palindromic

# number by permuting digits of a number

def largestPalindrome(s):

# string length

l = len(s)

# map that marks the occurrence of a number

m = defaultdict(lambda:0)

for i in range(0, l):

m[int(s[i])] += 1

# check the possibility of a

# palindromic number

if possibility(m, l, s) == False:

print(“Palindrome cannot be formed”)

return

# string array that stores the largest

# permuted palindromic number

largest = [None] * l

# pointer of front

front = 0

# greedily start from 9 to 0 and place the

# greater number in front and odd in the middle

for i in range(9, -1, -1):

# if the occurrence of number is odd

if m[i] & 1:

# place one odd occurring number

# in the middle

largest[l // 2] = chr(i + 48)

# decrease the count

m[i] -= 1

# place the rest of numbers greedily

while m[i] > 0:

largest[front] = chr(i + 48)

largest[l – front – 1] = chr(i + 48)

m[i] -= 2

front += 1

else:

# if all numbers occur even times,

# then place greedily

while m[i] > 0:

# place greedily at front

largest[front] = chr(i + 48)

largest[l – front – 1] = chr(i + 48)

# 2 numbers are placed,

# so decrease the count

m[i] -= 2

# increase placing position

front += 1

# print the largest string thus formed

for i in range(0, l):

print(largest[i], end = “”)

# Driver Code

if __name__ == “__main__”:

s = “313551”

largestPalindrome(s)

# This code is contributed by Rituraj Jain

**Output:**

531135

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

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