Given a number n, we can divide it in only three parts n/2, n/3 and n/4 (we will consider only integer part). The task is to find the maximum sum we can make by dividing number in three parts recursively and summing up them together.

Examples:

Input : n = 12 Output : 13 // We break n = 12 in three parts {12/2, 12/3, 12/4} // = {6, 4, 3}, now current sum is = (6 + 4 + 3) = 13 // again we break 6 = {6/2, 6/3, 6/4} = {3, 2, 1} = 3 + // 2 + 1 = 6 and further breaking 3, 2 and 1 we get maximum // summation as 1, so breaking 6 in three parts produces // maximum sum 6 only similarly breaking 4 in three // parts we can get maximum sum 4 and same for 3 also. // Thus maximum sum by breaking number in parts is=13 Input : n = 24 Output : 27 // We break n = 24 in three parts {24/2, 24/3, 24/4} // = {12, 8, 6}, now current sum is = (12 + 8 + 6) = 16 // As seen in example, recursively breaking 12 would // produce value 13. So our maximum sum is 13 + 8 + 6 = 27. // Note that recursively breaking 8 and 6 doesn't produce // more values, that is why they are not broken further. Input : n = 23 Output : 23 // we break n = 23 in three parts {23/2, 23/3, 23/4} = // {10, 7, 5}, now current sum is = (10 + 7 + 5) = 22. // Since after further breaking we can not get maximum // sum hence number is itself maximum i.e; answer is 23

A **simple solution** for this problem is to do it **recursively**. In each call we have to check only **max((max(n/2) + max(n/3) + max(n/4)), n)** and return it. Because either we can get maximum sum by breaking number in parts or number is itself maximum. Below is the implementation of recursive algorithm.

## C++

// A simple recursive C++ program to find // maximum sum by recursively breaking a // number in 3 parts. #include<bits/stdc++.h> using namespace std; // Function to find the maximum sum int breakSum(int n) { // base conditions if (n==0 || n == 1) return n; // recursively break the number and return // what maximum you can get return max((breakSum(n/2) + breakSum(n/3) + breakSum(n/4)), n); } // Driver program to run the case int main() { int n = 12; cout << breakSum(n); return 0; }

## Java

// A simple recursive JAVA program to find // maximum sum by recursively breaking a // number in 3 parts. import java.io.*; class GFG { // Function to find the maximum sum static int breakSum(int n) { // base conditions if (n==0 || n == 1) return n; // recursively break the number and return // what maximum you can get return Math.max((breakSum(n/2) + breakSum(n/3) + breakSum(n/4)), n); } // Driver program to test the above function public static void main (String[] args) { int n = 12; System.out.println(breakSum(n)); } } // This code is contributed by Amit Kumar

Output :

13

An **efficient solution** for this problem is to use **Dynamic programming** because while breaking the number in parts recursively we have to perform some overlapping problems. For example part of n = 30 will be {15,10,7} and part of 15 will be {7,5,3} so we have to repeat the process for 7 two times for further breaking. To avoid this overlapping problem we are using **Dynamic programming**. We store values in an array and if for any number in recursive calls we have already solution for that number currently so we diretly extract it from array.

## C++

// A Dynamic programming based C++ program // to find maximum sum by recursively breaking // a number in 3 parts. #include<bits/stdc++.h> #define MAX 1000000 using namespace std; int breakSum(int n) { int dp[n+1]; // base conditions dp[0] = 0, dp[1] = 1; // Fill in bottom-up manner using recursive // formula. for (int i=2; i<=n; i++) dp[i] = max(dp[i/2] + dp[i/3] + dp[i/4], i); return dp[n]; } // Driver program to run the case int main() { int n = 24; cout << breakSum(n); return 0; }

## Java

// A simple recursive JAVA program to find // maximum sum by recursively breaking a // number in 3 parts. import java.io.*; class GFG { final int MAX = 1000000; // Function to find the maximum sum static int breakSum(int n) { int dp[] = new int[n+1]; // base conditions dp[0] = 0; dp[1] = 1; // Fill in bottom-up manner using recursive // formula. for (int i=2; i<=n; i++) dp[i] = Math.max(dp[i/2] + dp[i/3] + dp[i/4], i); return dp[n]; } // Driver program to test the above function public static void main (String[] args) { int n = 24; System.out.println(breakSum(n)); } } // This code is contributed by Amit Kumar

Output:

13

Time Complexity : O(n)

Auxiliary Space : O(n)

This article is contributed by **Shashank Mishra ( Gullu )**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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