# Minimum number of palindromes required to express N as a sum | Set 1

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

11 is itself a palindrome.

Input:N = 65

Output:3

65 can be expressed as a sum of three palindromes (55, 9, 1).

**Approach:**

We can use Dynamic Programming to solve this problem. The idea is to first generate all the palindromes up to N in a sorted fashion, and then we can treat this problem as a variation of the subset sum problem, where we have to find the size of the smallest subset such that its sum is N.

**Below is the implementation of above approach:**

`// C++ implementation of above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Declaring the DP table as global variable ` `vector<vector<` `long` `long` `> > dp; ` ` ` ` ` `// 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 elements in a sorted ` `// array A[i..j] such that their sum is N ` `long` `long` `minimumSubsetSize(vector<` `int` `>& A, ` `int` `i, ` `int` `j, ` `int` `N) ` `{ ` ` ` `if` `(!N) ` ` ` `return` `0; ` ` ` ` ` `if` `(i > j || A[i] > N) ` ` ` `return` `INT_MAX; ` ` ` ` ` `if` `(dp[i][N]) ` ` ` `return` `dp[i][N]; ` ` ` ` ` `dp[i][N] = min(1 + minimumSubsetSize(A, i + 1, j, ` ` ` `N - A[i]), ` ` ` `minimumSubsetSize(A, i + 1, j, N)); ` ` ` ` ` `return` `dp[i][N]; ` `} ` ` ` `// Function to find the minimum ` `// number of palindromes that N ` `// can be expressed as a sum of ` `int` `minimumNoOfPalindromes(` `int` `N) ` `{ ` ` ` `// Getting the list of all palindromes upto N ` ` ` `vector<` `int` `> palindromes = generatePalindromes(N); ` ` ` ` ` `// Sorting the list of palindromes ` ` ` `sort(palindromes.begin(), palindromes.end()); ` ` ` ` ` `// Initializing the DP table ` ` ` `dp = vector<vector<` `long` `long` `> >(palindromes.size(), ` ` ` `vector<` `long` `long` `>(N + 1, 0)); ` ` ` ` ` `// Returning the required value ` ` ` `return` `minimumSubsetSize(palindromes, 0, ` ` ` `palindromes.size() - 1, N); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 65; ` ` ` `cout << minimumNoOfPalindromes(N); ` ` ` `return` `0; ` `} ` |

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**Output:**

3

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