# Recursively break a number in 3 parts to get maximum sum

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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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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