# 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 palindromeint 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 palindromesvector 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 sumint 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 codeint 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 sumimport 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 palindromedef 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 palindromesdef 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 sumdef 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 codeif __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 sumusing System;using System.Collections.Generic;class GFG{ // A utility for creating palindromestatic 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 palindromesstatic 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 sumstatic 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 codepublic 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(√(N)log N).

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