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# Minimum number of palindromes required to express N as a sum | Set 2

• Last Updated : 25 Aug, 2021

Given a number N, we have to find the minimum number of palindromes required to express N as a sum of them.
Examples

Input : N = 11
Output : 1
Explanation: 11 is itself a palindrome.
Input : N = 65
Output : 3
Explanation: 65 can be expressed as a sum of three palindromes (55, 9, 1).

In the previous post, we discussed a dynamic programming approach to this problem which had a time and space complexity of O(N3/2).
Cilleruelo, Luca, and Baxter proved in a 2016 research paper that every number can be expressed as the sum of a maximum of three palindromes in any base b >= 5 (this lower bound was later improved to 3). For the proof of this theorem, please refer to the original paper. We can make the use of this theorem by safely assuming the answer to be three if the number N is not itself a palindrome and cannot be expressed as the sum of two palindromes.
Below is the implementation of the above approach:

## C++

 // C++ program to find the minimum number of // palindromes required to express N as a sum   #include using namespace std;   // A utility for creating palindrome int createPalindrome(int input, bool isOdd) {     int n = input;     int palin = input;       // checks if number of digits is odd or even     // if odd then neglect the last digit of input in     // finding reverse as in case of odd number of     // digits middle element occur once     if (isOdd)         n /= 10;       // Creates palindrome by just appending reverse     // of number to itself     while (n > 0) {         palin = palin * 10 + (n % 10);         n /= 10;     }           return palin; }   // Function to generate palindromes vector generatePalindromes(int N) {     vector palindromes;     int number;       // Run two times for odd and even     // length palindromes     for (int j = 0; j < 2; j++) {           // Creates palindrome numbers with first half as i.         // Value of j decides whether we need an odd length         // or even length palindrome.         int i = 1;         while ((number = createPalindrome(i++, j)) <= N)             palindromes.push_back(number);     }       return palindromes; }   // Function to find the minimum // number of palindromes required // to express N as a sum int minimumNoOfPalindromes(int N) {     // Checking if the number is a palindrome     string a, b = a = to_string(N);     reverse(b.begin(), b.end());     if (a == b)         return 1;       // Checking if the number is a     // sum of two palindromes       // Getting the list of all palindromes upto N     vector palindromes = generatePalindromes(N);       // Sorting the list of palindromes     sort(palindromes.begin(), palindromes.end());       int l = 0, r = palindromes.size() - 1;     while (l < r) {         if (palindromes[l] + palindromes[r] == N)             return 2;         else if (palindromes[l] + palindromes[r] < N)             ++l;         else             --r;     }       // The answer is three if the     // control reaches till this point     return 3; }   // Driver code int main() {     int N = 65;           cout << minimumNoOfPalindromes(N);           return 0; }

## Java

 // Java program to find the minimum number of // palindromes required to express N as a sum import java.util.*;   class GFG {       // A utility for creating palindrome     static int createPalindrome(int input, int isOdd)     {         int n = input;         int palin = input;           // checks if number of digits is odd or even         // if odd then neglect the last digit of input in         // finding reverse as in case of odd number of         // digits middle element occur once         if (isOdd % 2 == 1)         {             n /= 10;         }           // Creates palindrome by just appending reverse         // of number to itself         while (n > 0)         {             palin = palin * 10 + (n % 10);             n /= 10;         }           return palin;     }       // Function to generate palindromes     static Vector generatePalindromes(int N)     {         Vector palindromes = new Vector<>();         int number;           // Run two times for odd and even         // length palindromes         for (int j = 0; j < 2; j++)         {               // Creates palindrome numbers with first half as i.             // Value of j decides whether we need an odd length             // or even length palindrome.             int i = 1;             while ((number = createPalindrome(i++, j)) <= N)             {                 palindromes.add(number);             }         }           return palindromes;     }       static String reverse(String input)     {         char[] temparray = input.toCharArray();         int left, right = 0;         right = temparray.length - 1;           for (left = 0; left < right; left++, right--)         {             // Swap values of left and right             char temp = temparray[left];             temparray[left] = temparray[right];             temparray[right] = temp;         }         return String.valueOf(temparray);     }       // Function to find the minimum     // number of palindromes required     // to express N as a sum     static int minimumNoOfPalindromes(int N)     {         // Checking if the number is a palindrome         String a = String.valueOf(N);         String b = String.valueOf(N);         b = reverse(b);         if (a.equals(b))         {             return 1;         }           // Checking if the number is a         // sum of two palindromes         // Getting the list of all palindromes upto N         Vector palindromes = generatePalindromes(N);           // Sorting the list of palindromes         Collections.sort(palindromes);           int l = 0, r = palindromes.size() - 1;         while (l < r)         {             if (palindromes.get(l) + palindromes.get(r) == N)             {                 return 2;             }             else if (palindromes.get(l) + palindromes.get(r) < N)             {                 ++l;             }             else             {                 --r;             }         }       // The answer is three if the         // control reaches till this point         return 3;     }       // Driver code     public static void main(String[] args)     {         int N = 65;         System.out.println(minimumNoOfPalindromes(N));     } }   // This code is contributed by Rajput-Ji

## Python3

 # Python3 program to find the minimum number of # palindromes required to express N as a sum   # A utility for creating palindrome def createPalindrome(_input, isOdd):        n = palin = _input       # checks if number of digits is odd or even     # if odd then neglect the last digit of _input in     # finding reverse as in case of odd number of     # digits middle element occur once     if isOdd:         n //= 10       # Creates palindrome by just appending reverse     # of number to itself     while n > 0:          palin = palin * 10 + (n % 10)         n //= 10           return palin    # Function to generate palindromes def generatePalindromes(N):        palindromes = []       # Run two times for odd and even     # length palindromes     for j in range(0, 2):            # Creates palindrome numbers with first half as i.         # Value of j decides whether we need an odd length         # or even length palindrome.         i = 1         number = createPalindrome(i, j)         while number <= N:             palindromes.append(number)             i += 1             number = createPalindrome(i, j)            return palindromes    # Function to find the minimum # number of palindromes required # to express N as a sum def minimumNoOfPalindromes(N):        # Checking if the number is a palindrome     b = a = str(N)     b = b[::-1]     if a == b:         return 1       # Checking if the number is a     # sum of two palindromes       # Getting the list of all palindromes upto N     palindromes = generatePalindromes(N)       # Sorting the list of palindromes     palindromes.sort()       l, r = 0, len(palindromes) - 1     while l < r:                   if palindromes[l] + palindromes[r] == N:             return 2         elif palindromes[l] + palindromes[r] < N:             l += 1         else:             r -= 1            # The answer is three if the     # control reaches till this point     return 3    # Driver code if __name__ == "__main__":        N = 65     print(minimumNoOfPalindromes(N))       # This code is contributed by Rituraj Jain

## C#

 // C# program to find the // minimum number of palindromes // required to express N as a sum using System; using System.Collections.Generic; class GFG{   // A utility for creating palindrome static int createPalindrome(int input,                             int isOdd) {   int n = input;   int palin = input;     // checks if number of digits   // is odd or even if odd then   // neglect the last digit of   // input in finding reverse   // as in case of odd number of   // digits middle element occur once   if (isOdd % 2 == 1)   {     n /= 10;   }     // Creates palindrome by   // just appending reverse   // of number to itself   while (n > 0)   {     palin = palin * 10 + (n % 10);     n /= 10;   }     return palin; }   // Function to generate palindromes static List generatePalindromes(int N) {   List palindromes = new List();   int number;     // Run two times for   // odd and even length   // palindromes   for (int j = 0; j < 2; j++)   {     // Creates palindrome numbers     // with first half as i. Value     // of j decides whether we need     // an odd length or even length     // palindrome.     int i = 1;     while ((number = createPalindrome(i++,                                       j)) <= N)     {       palindromes.Add(number);     }   }     return palindromes; }   static String reverse(String input) {   char[] temparray = input.ToCharArray();   int left, right = 0;   right = temparray.Length - 1;     for (left = 0;        left < right; left++, right--)   {     // Swap values of left and right     char temp = temparray[left];     temparray[left] = temparray[right];     temparray[right] = temp;   }       return String.Join("", temparray); }   // Function to find the minimum // number of palindromes required // to express N as a sum static int minimumNoOfPalindromes(int N) {   // Checking if the number   // is a palindrome   String a = String.Join("", N);   String b = String.Join("", N);   b = reverse(b);       if (a.Equals(b))   {     return 1;   }     // Checking if the number is   // a sum of two palindromes   // Getting the list of all   // palindromes upto N   List palindromes =             generatePalindromes(N);     // Sorting the list   // of palindromes   palindromes.Sort();     int l = 0,       r = palindromes.Count - 1;       while (l < r)   {     if (palindromes[l] +         palindromes[r] == N)     {       return 2;     }     else if (palindromes[l] +              palindromes[r] < N)     {       ++l;     }     else     {       --r;     }   }     // The answer is three if the   // control reaches till this point   return 3; }   // Driver code public static void Main(String[] args) {   int N = 65;   Console.WriteLine(minimumNoOfPalindromes(N)); } }   // This code is contributed by Rajput-Ji

## Javascript



Output:

3

Time Complexity: O(âˆš(N)log N).

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