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# 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) = 26
// 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 Practice

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``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 = 120;``    ``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`

## Python3

 `# A simple recursive Python3 program to ``# find maximum sum by recursively breaking  ``# a number in 3 parts. `` ` `# Function to find the maximum sum ``def` `breakSum(n) :`` ` `    ``# base conditions ``    ``if` `(n ``=``=` `0` `or` `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 Code``if` `__name__ ``=``=` `"__main__"` `: `` ` `    ``n ``=` `12``    ``print``(breakSum(n))`` ` `# This code is contributed by Ryuga`

## C#

 `// A simple recursive C# program to find``// maximum sum by recursively breaking a``// number in 3 parts.``using` `System;`` ` `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 () ``    ``{``        ``int` `n = 12;``         ` `        ``Console.WriteLine(breakSum(n));``    ``}``}`` ` `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output

`144`

Memoization of Recursive Solution

In the above solution, we see that there are many overlapping functions calls become so to reduce that we memoize the above recursive solution so it’s time complexity reduce exponentially to linear.

## C++

 `// memoization C++ program to find maximum sum by recursively breaking a number in 3 parts.`` ` `#include``using` `namespace` `std;`` ` `// Function to find the maximum sum``int` `breakSum(``int` `n, vector<``int``>&dp)``{``    ``// base conditions``    ``if` `(n == 0 || n == 1) {``        ``return` `n;``    ``}`` ` `    ``// if dp[n] already calculated then directly return so no overlapping problem calls``    ``if` `(dp[n] != -1) {``        ``return` `dp[n];``    ``}`` ` `    ``// break the number and return what maximum you can get``    ``return` `dp[n] = max((breakSum(n / 2, dp) + breakSum(n / 3, dp) + breakSum(n / 4, dp)),  n);``}`` ` `// Driver program to run the case``int` `main() {``    ``int` `n = 24;`` ` `    ``// dp array todo memoization``    ``vector<``int``>dp(n + 1, -1);`` ` `    ``cout << breakSum(n, dp);``    ``return` `0;``}`

## Java

 `// memoization Java program to find maximum sum by recursively breaking a number in 3 parts.`` ` `class` `Main {`` ` `    ``// Function to find the maximum sum``    ``public` `static` `int` `breakSum(``int` `n, ``int``[] dp)``    ``{``        ``// base conditions``        ``if` `(n == ``0` `|| n == ``1``)``            ``return` `n;`` ` `        ``// if dp[n] already calculated then directly return so no overlapping problem calls``        ``if` `(dp[n] != -``1``) {``            ``return` `dp[n];``        ``}`` ` `        ``// break the number and return what maximum you can get``        ``return` `dp[n] = Math.max((breakSum(n / ``2``, dp) + breakSum(n / ``3``, dp) + breakSum(n / ``4``, dp)),  n);`` ` `    ``}`` ` `    ``// Driver program to test the above function``    ``public` `static` `void` `main (String[] args) {``        ``int` `n = ``24``;``        ``int``[] dp = ``new` `int``[n+``1``];``        ``for` `(``int` `i = ``0``; i <= n; i++) dp[i] = -``1``;``        ``System.out.println(breakSum(n, dp));``    ``}``}`` ` `// This code is contributed by ajaymakvana.`

## Python3

 `# Python3 code to find maximum sum by recursively breaking a number in 3 parts.`` ` `# Function to find the maximum sum``def` `breakSum(n, dp):``   ` `    ``# base conditions``    ``if` `n ``=``=` `0` `or` `n ``=``=` `1``:``        ``return` `n``       ` `    ``# if dp[n] already calculated then directly return so no overlapping problem calls``    ``if` `dp[n] !``=` `-``1``:``        ``return` `dp[n]``       ` `    ``# break the number and return what maximum you can get``    ``dp[n] ``=` `max``((breakSum(n ``/``/` `2``, dp) ``+` `breakSum(n ``/``/` `3``, dp) ``+` `breakSum(n ``/``/` `4``, dp)), n)``    ``return` `dp[n]`` ` `# Driver program to run the case``if` `__name__ ``=``=` `"__main__"``:``    ``n ``=` `24``     ` `    ``# dp array todo memoization``    ``dp ``=` `[``-``1` `for` `i ``in` `range``(n ``+` `1``)]``    ``print``(breakSum(n, dp))`` ` `    ``# This code is contributed by lokehpotta20.`

## C#

 `// memoization C# program to find maximum sum by ``// recursively breaking a number in 3 parts.``using` `System;``using` `System.Collections.Generic;`` ` `class` `GFG {`` ` `  ``// Function to find the maximum sum``  ``static` `int` `breakSum(``int` `n, List<``int``> dp)``  ``{``    ``// base conditions``    ``if` `(n == 0 || n == 1) {``      ``return` `n;``    ``}`` ` `    ``// if dp[n] already calculated then directly ``    ``// return so no overlapping problem calls``    ``if` `(dp[n] != -1) {``      ``return` `dp[n];``    ``}`` ` `    ``// break the number and return what maximum you can get``    ``return` `dp[n] = Math.Max((breakSum(n / 2, dp) + ``                             ``breakSum(n / 3, dp) + ``                             ``breakSum(n / 4, dp)),  n);``  ``}`` ` `  ``// Driver program to run the case``  ``static` `public` `void` `Main()``  ``{``    ``int` `n = 24;`` ` `    ``// dp array todo memoization``    ``List<``int``>dp = ``new` `List<``int``>();``    ``for``(``int` `i = 0; i < n + 1; i++)``      ``dp.Add(-1);`` ` `    ``Console.Write(breakSum(n, dp));``  ``}``}`` ` `// This code is contributed by ratiagrawal.`

## Javascript

 `// memoization Javascript program to ``// find maximum sum by recursively breaking a number in 3 parts.`` ` `// Function to find the maximum sum``function` `breakSum(n, dp)``{``    ``// base conditions``    ``if` `(n == 0 || n == 1) {``        ``return` `n;``    ``}`` ` `    ``// if dp[n] already calculated then directly return ``    ``// so no overlapping problem calls``    ``if` `(dp[n] != -1) {``        ``return` `dp[n];``    ``}`` ` `    ``// break the number and return what maximum you can get``    ``return` `dp[n] = Math.max((breakSum(parseInt(n / 2), dp) + ``           ``breakSum(parseInt(n / 3), dp) + breakSum(parseInt(n / 4), dp)),  n);`` ` `}`` ` `// Driver program to run the case``let n = 24;`` ` `// dp array todo memoization``let dp=``new` `Array(n + 1).fill(-1);`` ` `console.log(breakSum(n, dp));`

Output

`27`

Time Complexity: O(n)
Auxiliary Space: O(n) (as we are creating dp array of size n+1)

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 directly extract it from array.

## C++

 `// A Dynamic programming based C++ program``// to find maximum sum by recursively breaking``// a number in 3 parts.``#include``#define MAX 1000000``using` `namespace` `std;`` ` `int` `breakSum(``int` `n)``{``    ``int` `dp[n+1];`` ` `    ``// base conditions``    ``dp = 0, dp = 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`

## Python3

 `# A Dynamic programming based Python program``# to find maximum sum by recursively breaking``# a number in 3 parts.`` ` `MAX` `=` `1000000`` ` `def` `breakSum(n):`` ` `    ``dp ``=` `[``0``]``*``(n``+``1``)`` ` `    ``# base conditions``    ``dp[``0``] ``=` `0``    ``dp[``1``] ``=` `1`` ` `    ``# Fill in bottom-up manner using recursive``    ``# formula.``    ``for` `i ``in` `range``(``2``, n``+``1``):``        ``dp[i] ``=` `max``(dp[``int``(i``/``2``)] ``+` `dp[``int``(i``/``3``)] ``+` `dp[``int``(i``/``4``)], i);`` ` `    ``return` `dp[n]`` ` ` ` `# Driver program to run the case``n ``=` `24``print``(breakSum(n))`` ` `# This code is contributed by``# Smitha Dinesh Semwal`

## C#

 `// A simple recursive C# program to find``// maximum sum by recursively breaking a``// number in 3 parts.``using` `System;`` ` `class` `GFG {``     ` `    ``// Function to find the maximum sum``    ``static` `int` `breakSum(``int` `n)``    ``{``        ``int` `[]dp = ``new` `int``[n+1];`` ` `        ``// base conditions``        ``dp = 0; dp = 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 () ``    ``{``        ``int` `n = 24;``        ``Console.WriteLine(breakSum(n));``    ``}``}`` ` `// This code is contributed by anuj_67.`

## PHP

 ``

## Javascript

 ``

Output

`27`

Time Complexity: O(n)
Auxiliary Space: O(n)

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