# Find the maximum number of composite summands of a number

Given an integer N(1<=N<=10^9). The task is to represent N as a sum of the maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. There can be multiple queries

Examples:

```Input : 12
Output : 3
Explanation : 12 can be written has 4 + 4 + 4 or 6 + 6 or 8 + 4
But, 4 + 4 + 4 has maximum number of summands.

Input : 7
Output : -1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : Note that minimal composite number is equal to 4. So it is quite logical that there will be a lot of 4 in a splitting of big numbers. Let’s write for small numbers (1<=M<=N) dpN be the number of composite summands in splitting of N.

Let’s find an answer for all numbers from 1 to 15. Several observations:

1. Only 4, 6, 9 occurs in optimal splittings.
2. It is not beneficial to use 6 or 9 more than once because 6 + 6 = 4 + 4 + 4, 9 + 9 = 6 + 6 + 6.
3. 12, 13, 14, 15 have valid splittings.

Let’s prove that all numbers that are greater than 15 will have 4 in optimal splitting. Let’s guess that it is incorrect. If the minimal number in splitting is neither 4 nor 6 nor 9 then this number will have some non-trivial splitting by induction.

If this number either 6 or 9 and we will decrease query by this number then we will sooner or later get some small number (which is less or equal than 15). There is no splitting of small numbers or it contains 4 in splitting (and it contradicts with minimality of the first number) or it contains 6 and 9. So we have contradiction in all cases.

We can subtract 4 from any big query and our solution is correct.

If our query n is small number let’s print dpn. Else let’s find minimal number k such that n – 4·k is a small number. Then print k + dpn – 4·k.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `const` `int` `maxn = 16; ` ` `  `// Function to generate the dp array ` `vector<``int``> precompute() ` `{ ` `    ``vector<``int``> dp(maxn, -1); ` `    ``dp = 0; ` ` `  `    ``for` `(``int` `i = 1; i < maxn; ++i) { ` ` `  `        ``// combination of three integers ` `        ``for` `(``auto` `j : vector<``int``>{ 4, 6, 9 }) { ` ` `  `            ``// take the maxium number of summands ` `            ``if` `(i >= j && dp[i - j] != -1) { ` `                ``dp[i] = max(dp[i], dp[i - j] + 1); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `dp; ` `} ` ` `  `// Function to find the maximum number of summands ` `int` `Maximum_Summands(vector<``int``> dp, ``int` `n) ` `{ ` `    ``// If n is a smaller number, less than 16 ` `    ``// return dp[n] ` `    ``if` `(n < maxn) ` `        ``return` `dp[n]; ` ` `  `    ``else` `{ ` ` `  `        ``// Else, find a minimal number t ` `        ``// as explained in solution ` `        ``int` `t = (n - maxn) / 4 + 1; ` `        ``return` `t + dp[n - 4 * t]; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 12; ` ` `  `    ``// Generate dp array ` `    ``vector<``int``> dp = precompute(); ` ` `  `    ``cout << Maximum_Summands(dp, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG ` `{ ` ` `  `static` `int` `maxn = ``16``;  ` ` `  `// Function to generate the dp array  ` `static` `int``[] precompute()  ` `{  ` `int` `dp[] = ``new` `int``[maxn], arr[]={ ``4``, ``6``, ``9` `}; ` ` `  `// initilize ` `for``(``int` `i = ``0``; i < maxn; i++)dp[i] = -``1``; ` ` `  `dp[``0``] = ``0``;  ` ` `  `for` `(``int` `i = ``1``; i < maxn; ++i)  ` `{  ` ` `  `    ``// combination of three integers  ` `    ``for` `(``int` `k = ``0``; k < ``3``; k++)  ` `    ``{ ` `        ``int` `j = arr[k]; ` ` `  `        ``// take the maxium number of summands  ` `        ``if` `(i >= j && dp[i - j] != -``1``)  ` `        ``{  ` `            ``dp[i] = Math.max(dp[i], dp[i - j] + ``1``);  ` `        ``}  ` `    ``}  ` `}  ` ` `  `return` `dp;  ` `}  ` ` `  `// Function to find the maximum number of summands  ` `static` `int` `Maximum_Summands(``int``[] dp, ``int` `n)  ` `{  ` `// If n is a smaller number, less than 16  ` `// return dp[n]  ` `if` `(n < maxn)  ` `    ``return` `dp[n];  ` ` `  `else` `{  ` ` `  `    ``// Else, find a minimal number t  ` `    ``// as explained in solution  ` `    ``int` `t = (n - maxn) / ``4` `+ ``1``;  ` `    ``return` `t + dp[n - ``4` `* t];  ` `}  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[])  ` `{  ` `    ``int` `n = ``12``;  ` ` `  `    ``// Generate dp array  ` `    ``int``[] dp = precompute();  ` ` `  `    ``System.out.println(Maximum_Summands(dp, n));  ` `} ` `}  ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python 3 implementation of the above approach ` `global` `maxn ` `maxn ``=` `16` ` `  `# Function to generate the dp array ` `def` `precompute(): ` `    ``dp ``=` `[``-``1` `for` `i ``in` `range``(maxn)] ` `    ``dp[``0``] ``=` `0` ` `  `    ``v ``=` `[``4``, ``6``, ``9``] ` ` `  `    ``for` `i ``in` `range``(``1``, maxn, ``1``): ` `         `  `        ``# combination of three integers ` `        ``for` `k ``in` `range``(``3``): ` `            ``j ``=` `v[k] ` `             `  `            ``# take the maxium number of summands ` `            ``if` `(i >``=` `j ``and` `dp[i ``-` `j] !``=` `-``1``): ` `                ``dp[i] ``=` `max``(dp[i], dp[i ``-` `j] ``+` `1``) ` ` `  `    ``return` `dp ` ` `  `# Function to find the maximum number of summands ` `def` `Maximum_Summands(dp, n): ` `     `  `    ``# If n is a smaller number,  ` `    ``# less than 16, return dp[n] ` `    ``if` `(n < maxn): ` `        ``return` `dp[n] ` ` `  `    ``else``: ` `         `  `        ``# Else, find a minimal number t ` `        ``# as explained in solution ` `        ``t ``=` `int``((n ``-` `maxn) ``/` `4``)``+` `1` `        ``return` `t ``+` `dp[n ``-` `4` `*` `t] ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `12` ` `  `    ``# Generate dp array ` `    ``dp ``=` `precompute() ` ` `  `    ``print``(Maximum_Summands(dp, n)) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the above approach ` `using` `System; ` `using` `System.Collections; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `maxn = 16; ` `static` `int``[] dp = ``new` `int``[maxn + 1]; ` ` `  `// Function to generate the dp array ` `static` `void` `precompute() ` `{ ` `    ``for``(``int` `i = 0; i <= maxn; i++) ` `    ``dp[i] = -1; ` `    ``dp = 0; ` `    ``int``[] vec = { 4, 6, 9 }; ` `    ``for` `(``int` `i = 1; i < maxn; ++i)  ` `    ``{ ` ` `  `        ``// combination of three integers ` `        ``foreach` `(``int` `j ``in` `vec) ` `        ``{ ` ` `  `            ``// take the maxium number of summands ` `            ``if` `(i >= j && dp[i - j] != -1)  ` `            ``{ ` `                ``dp[i] = Math.Max(dp[i], dp[i - j] + 1); ` `            ``} ` `        ``} ` `    ``} ` ` `  `} ` ` `  `// Function to find the maximum number of summands ` `static` `int` `Maximum_Summands(``int` `n) ` `{ ` `    ``// If n is a smaller number, less than 16 ` `    ``// return dp[n] ` `    ``if` `(n < maxn) ` `        ``return` `dp[n]; ` ` `  `    ``else`  `    ``{ ` ` `  `        ``// Else, find a minimal number t ` `        ``// as explained in solution ` `        ``int` `t = (n - maxn) / 4 + 1; ` `        ``return` `t + dp[n - 4 * t]; ` `    ``} ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `n = 12; ` ` `  `    ``// Generate dp array ` `    ``precompute(); ` ` `  `    ``Console.WriteLine(Maximum_Summands(n)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

## PHP

 `= ``\$j` `&& ``\$dp``[``\$i` `- ``\$j``] != -1)  ` `            ``{  ` `                ``\$dp``[``\$i``] = max(``\$dp``[``\$i``],  ` `                              ``\$dp``[``\$i` `- ``\$j``] + 1);  ` `            ``}  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `\$dp``;  ` `}  ` ` `  `// Function to find the maximum  ` `// number of summands  ` `function` `Maximum_Summands(``\$dp``, ``\$n``)  ` `{  ` `    ``// If n is a smaller number,  ` `    ``// less than 16 return dp[n]  ` `    ``if` `(``\$n` `< ``\$GLOBALS``[``'maxn'``])  ` `        ``return` `\$dp``[``\$n``];  ` ` `  `    ``else` `    ``{  ` ` `  `        ``// Else, find a minimal number t  ` `        ``// as explained in solution  ` `        ``\$t` `= (``\$n` `- ``\$GLOBALS``[``'maxn'``]) / 4 + 1;  ` `        ``return` `\$t` `+ ``\$dp``[``\$n` `- 4 * ``\$t``];  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `\$n` `= 12;  ` ` `  `// Generate dp array  ` `\$dp` `= precompute();  ` ` `  `echo` `Maximum_Summands(``\$dp``, ``\$n``);  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```3
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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