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

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(N^3/2).
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 <bits/stdc++.h>
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<int> generatePalindromes(int N)
{
    vector<int> 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<int> 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<Integer> generatePalindromes(int N)
    {
        Vector<Integer> 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<Integer> 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<int> generatePalindromes(int N)
{
  List<int> palindromes = new List<int>();
  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<int> 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

<script>

// Javascript program to find the
// minimum number of palindromes
// required to express N as a sum

// A utility for creating palindrome
function createPalindrome(input, isOdd)
{
  var n = input;
  var 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 = parseInt(n/10);
  }

  // Creates palindrome by
  // just appending reverse
  // of number to itself
  while (n > 0)
  {
    palin = palin * 10 + (n % 10);
    n = parseInt(n/10);
  }

  return palin;
}

// Function to generate palindromes
function generatePalindromes(N)
{
  var palindromes = [];
  var number;

  // Run two times for
  // odd and even length
  // palindromes
  for (var 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.
    var i = 1;
    while ((number = createPalindrome(i++,
                                      j)) <= N)
    {
      palindromes.push(number);
    }
  }

  return palindromes;
}

function reverse(input)
{
  var temparray = input.split('');
  var left, right = 0;
  right = temparray.length - 1;

  for (left = 0;
       left < right; left++, right--)
  {
    // Swap values of left and right
    var temp = temparray[left];
    temparray[left] = temparray[right];
    temparray[right] = temp;
  }

  return temparray.join('');
}

// Function to find the minimum
// number of palindromes required
// to express N as a sum
function minimumNoOfPalindromes(N)
{
  // Checking if the number
  // is a palindrome
  var a = N.toString();
  var b = N.toString();
  b = reverse(b);

  if (a == b)
  {
    return 1;
  }

  // Checking if the number is
  // a sum of two palindromes
  // Getting the list of all
  // palindromes upto N
  var palindromes =
            generatePalindromes(N);

  // Sorting the list
  // of palindromes
  palindromes.sort();

  var l = 0,
      r = palindromes.length - 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
var N = 65;
document.write(minimumNoOfPalindromes(N));

// This code is contributed by rrrtnx.
</script>

Output:

3

Time Complexity: O(√(N)log N).