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(N3/2).

Cilleruelo, Luca, and Baxter proved in a 2016 research paper that every number can be expressed as the sum of maximum 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++

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// 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 revers
    // 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;
}

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Java

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// 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 revers
        // 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

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Python3

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# 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 revers 
    # 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

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Output:

3

Time Complexity: O(√(N)log N).



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Improved By : rituraj_jain, Rajput-Ji