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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.#includeusing namespace std;  // Function to find the maximum sumint 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 caseint 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 Codeif __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.



## 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.  #includeusing namespace std;  // Function to find the maximum sumint breakSum(int n, vector&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 caseint main() {    int n = 24;      // dp array todo memoization    vectordp(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 sumdef 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 caseif __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 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    Listdp = new List();    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 sumfunction 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 caselet n = 24;  // dp array todo memoizationlet 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 1000000using 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 caseint 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 casen = 24print(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] = 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 ()     {        int n = 24;        Console.WriteLine(breakSum(n));    }}  // This code is contributed by anuj_67.



## Javascript



Output

27

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

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