# Maximum Product Cutting | DP-36

• Difficulty Level : Medium
• Last Updated : 21 Jun, 2022

Given a rope of length n meters, cut the rope in different parts of integer lengths in a way that maximizes product of lengths of all parts. You must make at least one cut. Assume that the length of rope is more than 2 meters.

Examples:

Input: n = 2
Output: 1 (Maximum obtainable product is 1*1)

Input: n = 3
Output: 2 (Maximum obtainable product is 1*2)

Input: n = 4
Output: 4 (Maximum obtainable product is 2*2)

Input: n = 5
Output: 6 (Maximum obtainable product is 2*3)

Input: n = 10
Output: 36 (Maximum obtainable product is 3*3*4)

1) Optimal Substructure:

This problem is similar to Rod Cutting Problem. We can get the maximum product by making a cut at different positions and comparing the values obtained after a cut. We can recursively call the same function for a piece obtained after a cut.
Let maxProd(n) be the maximum product for a rope of length n. maxProd(n) can be written as following.
maxProd(n) = max(i*(n-i), maxProdRec(n-i)*i) for all i in {1, 2, 3 .. n}

2) Overlapping Subproblems:

Following is simple recursive implementation of the problem. The implementation simply follows the recursive structure mentioned above.

## C++

 // A Naive Recursive method to find maximum product#include using namespace std; // Utility function to get the maximum of two and three integersint max(int a, int b) { return (a > b)? a : b;}int max(int a, int b, int c) { return max(a, max(b, c));} // The main function that returns maximum product obtainable// from a rope of length nint maxProd(int n){    // Base cases    if (n == 0 || n == 1) return 0;     // Make a cut at different places and take the maximum of all    int max_val = 0;    for (int i = 1; i < n; i++)      max_val = max(max_val, i*(n-i), maxProd(n-i)*i);     // Return the maximum of all values    return max_val;} /* Driver program to test above functions */int main(){    cout << "Maximum Product is " << maxProd(10);    return 0;}

## Java

 // Java program to find maximum productimport java.io.*; class GFG {     // The main function that returns    // maximum product obtainable from    // a rope of length n    static int maxProd(int n)    {        // Base cases        if (n == 0 || n == 1) return 0;         // Make a cut at different places        // and take the maximum of all        int max_val = 0;        for (int i = 1; i < n; i++)        max_val = Math.max(max_val,                  Math.max(i * (n - i),                   maxProd(n - i) * i));         // Return the maximum of all values        return max_val;    }       /* Driver program to test above functions */    public static void main(String[] args)    {        System.out.println("Maximum Product is "                            + maxProd(10));    }}// This code is contributed by Prerna Saini

## Python3

 # The main function that returns maximum# product obtainable from a rope of length n def maxProd(n):         # Base cases    if (n == 0 or n == 1):        return 0      # Make a cut at different places    # and take the maximum of all    max_val = 0    for i in range(1, n - 1):        max_val = max(max_val, max(i * (n - i), maxProd(n - i) * i))      #Return the maximum of all values    return max_val;   # Driver program to test above functionsprint("Maximum Product is ", maxProd(10));     # This code is contributed# by Sumit Sudhakar

## C#

 // C# program to find maximum productusing System; class GFG {         // The main function that returns     // the max possible product    static int maxProd(int n)    {         // n equals to 2 or 3 must        // be handled explicitly        if (n == 2 || n == 3)            return (n - 1);         // Keep removing parts of size        // 3 while n is greater than 4        int res = 1;        while (n > 4) {            n -= 3;             // Keep multiplying 3 to res            res *= 3;        }         // The last part multiplied        // by previous parts        return (n * res);    }     // Driver code    public static void Main()    {        Console.WriteLine("Maximum Product is "                                + maxProd(10));    }} // This code is contributed by Sam007



## Javascript



Output

Maximum Product is 36

Considering the above implementation, following is recursion tree for a Rope of length 5.

In the above partial recursion tree, mP(3) is being solved twice. We can see that there are many subproblems which are solved again and again. Since same subproblems are called again, this problem has Overlapping Subproblems property. So the problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array val[] in bottom up manner.

## C++

 // C++ code to implement the approach\ // A Dynamic Programming solution for Max Product Problemint maxProd(int n){   int val[n+1];   val[0] = val[1] = 0;     // Build the table val[] in bottom up manner and return   // the last entry from the table   for (int i = 1; i <= n; i++)   {      int max_val = 0;      for (int j = 1; j <= i; j++)         max_val = max(max_val, (i-j)*j, j*val[i-j]);      val[i] = max_val;   }   return val[n];} // This code is contributed by sanjoy_62.

## C

 // A Dynamic Programming solution for Max Product Problemint maxProd(int n){   int val[n+1];   val[0] = val[1] = 0;     // Build the table val[] in bottom up manner and return   // the last entry from the table   for (int i = 1; i <= n; i++)   {      int max_val = 0;      for (int j = 1; j <= i; j++)         max_val = max(max_val, (i-j)*j, j*val[i-j]);      val[i] = max_val;   }   return val[n];}

## Java

 // A Dynamic Programming solution for Max Product Problemint maxProd(int n){   int val[n+1];   val[0] = val[1] = 0;     // Build the table val[] in bottom up manner and return   // the last entry from the table   for (int i = 1; i <= n; i++)   {      int max_val = 0;      for (int j = 1; j <= i; j++)         max_val = Math.max(max_val, (i-j)*j, j*val[i-j]);      val[i] = max_val;   }   return val[n];} // This code is contributed by umadevi9616

## Python3

 # A Dynamic Programming solution for Max Product Problemdef maxProd(n):   val= [0 for i in range(n+1)];      # Build the table val in bottom up manner and return   # the last entry from the table   for i in range(1,n+1):      max_val = 0;      for j in range(1,i):         max_val = max(max_val, (i-j)*j, j*val[i-j]);      val[i] = max_val;    return val[n]; # This code is contributed by gauravrajput1

## C#

 // A Dynamic Programming solution for Max Product Problemint maxProd(int n){   int []val = new int[n+1];   val[0] = val[1] = 0;     // Build the table val[] in bottom up manner and return   // the last entry from the table   for (int i = 1; i <= n; i++)   {      int max_val = 0;      for (int j = 1; j <= i; j++)         max_val = Math.Max(max_val, (i-j)*j, j*val[i-j]);      val[i] = max_val;   }   return val[n];} // This code is contributed by umadevi9616

## Javascript



Time Complexity of the Dynamic Programming solution is O(n^2) and it requires O(n) extra space.

A Tricky Solution:

If we see some examples of this problems, we can easily observe following pattern.
The maximum product can be obtained be repeatedly cutting parts of size 3 while size is greater than 4, keeping the last part as size of 2 or 3 or 4. For example, n = 10, the maximum product is obtained by 3, 3, 4. For n = 11, the maximum product is obtained by 3, 3, 3, 2. Following is the implementation of this approach.

## C++

 #include using namespace std; /* The main function that returns the max possible product */int maxProd(int n){   // n equals to 2 or 3 must be handled explicitly   if (n == 2 || n == 3) return (n-1);    // Keep removing parts of size 3 while n is greater than 4   int res = 1;   while (n > 4)   {       n -= 3;       res *= 3; // Keep multiplying 3 to res   }   return (n * res); // The last part multiplied by previous parts} /* Driver program to test above functions */int main(){    cout << "Maximum Product is " << maxProd(10);    return 0;}

## Java

 // Java program to find maximum productimport java.io.*; class GFG {     /* The main function that returns the    max possible product */    static int maxProd(int n)    {         // n equals to 2 or 3 must be handled    // explicitly    if (n == 2 || n == 3) return (n-1);     // Keep removing parts of size 3    // while n is greater than 4    int res = 1;    while (n > 4)    {        n -= 3;                 // Keep multiplying 3 to res        res *= 3;    }         // The last part multiplied by    // previous parts    return (n * res);    }     /* Driver program to test above functions */    public static void main(String[] args)    {        System.out.println("Maximum Product is "                            + maxProd(10));    }  }// This code is contributed by Prerna Saini

## Python3

 # The main function that returns the# max possible product def maxProd(n):         # n equals to 2 or 3 must    # be handled explicitly    if (n == 2 or n == 3):        return (n - 1)      # Keep removing parts of size 3    # while n is greater than 4    res = 1    while (n > 4):        n -= 3;                  # Keep multiplying 3 to res        res *= 3;         # The last part multiplied    # by previous parts    return (n * res) # Driver program to test above functionsprint("Maximum Product is ", maxProd(10));     # This code is contributed# by Sumit Sudhakar

## C#

 // C# program to find maximum productusing System; class GFG {     // The main function that returns    // maximum product obtainable from    // a rope of length n    static int maxProd(int n)    {        // Base cases        if (n == 0 || n == 1)            return 0;         // Make a cut at different places        // and take the maximum of all        int max_val = 0;        for (int i = 1; i < n; i++)            max_val = Math.Max(max_val,                    Math.Max(i * (n - i),                     maxProd(n - i) * i));         // Return the maximum of all values        return max_val;    }     // Driver code    public static void Main()     {        Console.WriteLine("Maximum Product is "                                + maxProd(10));     }} // This code is contributed by Sam007

## PHP

 4){    \$n = \$n - 3;         // Keep multiplying 3 to res    \$res = \$res * 3;} // The last part multiplied// by previous partsreturn (\$n * \$res);} // Driver codeecho ("Maximum Product is ");echo(maxProd(10)); // This code is contributed// by Shivi_Aggarwal?>

## Javascript



Output

Maximum Product is 36

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