# Maximum Product Cutting | DP-36

Last Updated : 25 Apr, 2023

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 integers` `int` `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 n` `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 = 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 product` `import` `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 functions ` `print``(``"Maximum Product is "``, maxProd(``10``));` `    `  `# This code is contributed` `# by Sumit Sudhakar`

## C#

 `// C# program to find maximum product` `using` `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`

## PHP

 ``

## Javascript

 ``

Output

`Maximum Product is 36`

Time complexity: O(n^2)

The time complexity of the maxProd function is O(n^2), because it contains a loop that iterates n-1 times, and inside the loop it calls itself recursively with a smaller input size (n-i). The maximum recursion depth is n, so the total time complexity is n * (n-1) = O(n^2).

Space complexity: O(n)

The space complexity of the maxProd function is O(n), because the maximum recursion depth is n, and each recursive call adds a new activation record to the call stack, which contains local variables and return addresses. Therefore, the space used by the call stack is proportional to the input size n.

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 Problem` `int` `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 Problem` `int` `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 Problem` `int` `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 Problem` `def` `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 Problem` `int` `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 product` `import` `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 functions ` `print``(``"Maximum Product is "``, maxProd(``10``));` `    `  `# This code is contributed ` `# by Sumit Sudhakar`

## C#

 `// C# program to find maximum product` `using` `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 parts` `return` `(``\$n` `* ``\$res``); ` `}`   `// Driver code` `echo` `(``"Maximum Product is "``);` `echo``(maxProd(10));`   `// This code is contributed ` `// by Shivi_Aggarwal ` `?>`

## Javascript

 ``

Output

`Maximum Product is 36`

Time Complexity: O(n/3) ~= O(n), as here in every loop step we do decrement of 3 from n so it’s take n/3 – 1 iteration to make n less than or equal to 4 so ultimately we need O(n) time complexity.
Space Complexity: O(1), as we don’t need any extra space to generate answer

More Efficient Solution (log(n) time complexity solution)

In the above solution, we are using a while loop in which on every iteration of look we are subtracting 3 from n and multiplying the result with 3. But If we observe mathematically then, for any number n, we can subtract (n/3) times 3 from that number. so using this observation we can eliminate that while loop from the code and instead of that we directly do (n/3) times 3 subtraction from n. so after knowing that number we can use the pow() function to multiply the result which takes log(n) time complexity to multiply. here the while loop break condition is (n>4) so we need to handle that case by the if-else statement as shown in the code.

## C++

 `#include ` `using` `namespace` `std;`   `/* The main function that returns the max possible product` ` ``*/` `int` `maxProd(``int` `n)` `{` `    ``// edge case` `    ``if` `((n == 2) || (n == 3)) {` `        ``return` `n - 1;` `    ``}`   `    ``// how many times we can subtract 3 from n is (n/3)` `    ``int` `num_threes = n / 3;`   `    ``int` `num_twos = 0;`   `    ``// specifically for handling case where n = 4, 7, 10,` `    ``// ...` `    ``if` `(n % 3 == 1) {` `        ``num_threes -= 1;` `        ``num_twos = 2;` `    ``}` `    ``// specifically for handling case where n = 5, 8, 11,` `    ``// ...` `    ``else` `if` `(n % 3 == 2) {` `        ``num_twos = 1;` `    ``}`   `    ``int` `res = 1;`   `    ``res *= ``pow``(3, num_threes);`   `    ``res *= ``pow``(2, num_twos);`   `    ``return` `res;` `}`   `/* Driver program to test above functions */` `int` `main()` `{` `    ``cout << ``"Maximum Product is "` `<< maxProd(10);` `    ``return` `0;` `}`

## Java

 `// Java Program for the above approach` `import` `java.lang.Math;`   `public` `class` `Main {`   `    ``public` `static` `void` `main(String[] args) {` `        ``System.out.println(``"Maximum Product is "` `+ maxProd(``10``));` `    ``}`   `    ``/* The main function that returns the max possible product */` `    ``public` `static` `int` `maxProd(``int` `n) {` `        ``// edge case` `        ``if` `((n == ``2``) || (n == ``3``)) {` `            ``return` `n - ``1``;` `        ``}`   `        ``// how many times we can subtract 3 from n is (n/3)` `        ``int` `num_threes = n / ``3``;`   `        ``int` `num_twos = ``0``;`   `        ``// specifically for handling case where n = 4, 7, 10,` `        ``// ...` `        ``if` `(n % ``3` `== ``1``) {` `            ``num_threes -= ``1``;` `            ``num_twos = ``2``;` `        ``}` `        ``// specifically for handling case where n = 5, 8, 11,` `        ``// ...` `        ``else` `if` `(n % ``3` `== ``2``) {` `            ``num_twos = ``1``;` `        ``}`   `        ``int` `res = ``1``;`   `        ``res *= Math.pow(``3``, num_threes);`   `        ``res *= Math.pow(``2``, num_twos);`   `        ``return` `res;` `    ``}` `}`   `// Contributed by adityasharmadev01`

## Python3

 `import` `math`     `def` `max_prod(n):` `    ``"""` `    ``The main function that returns the max possible product` `    ``"""` `    ``# edge case` `    ``if` `n ``=``=` `2` `or` `n ``=``=` `3``:` `        ``return` `n ``-` `1`   `    ``# how many times we can subtract 3 from n is (n/3)` `    ``num_threes ``=` `n ``/``/` `3`   `    ``num_twos ``=` `0`   `    ``# specifically for handling case where n = 4, 7, 10, ...` `    ``if` `n ``%` `3` `=``=` `1``:` `        ``num_threes ``-``=` `1` `        ``num_twos ``=` `2` `    ``# specifically for handling case where n = 5, 8, 11, ...` `    ``elif` `n ``%` `3` `=``=` `2``:` `        ``num_twos ``=` `1`   `    ``res ``=` `1`   `    ``res ``*``=` `math.``pow``(``3``, num_threes)`   `    ``res ``*``=` `math.``pow``(``2``, num_twos)`   `    ``return` `int``(res)`     `# Driver program to test above function` `print``(``"Maximum Product is"``, max_prod(``10``))`

## C#

 `using` `System;` `public` `class` `Program {` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``Console.WriteLine(``"Maximum Product is "` `                          ``+ maxProd(10));` `    ``}`   `    ``/* The main function that returns the max possible` `     ``* product */` `    ``public` `static` `int` `maxProd(``int` `n)` `    ``{` `        ``// edge case` `        ``if` `((n == 2) || (n == 3)) {` `            ``return` `n - 1;` `        ``}`   `        ``// how many times we can subtract 3 from n is (n/3)` `        ``int` `num_threes = n / 3;`   `        ``int` `num_twos = 0;`   `        ``// specifically for handling case where n = 4, 7,` `        ``// 10,` `        ``// ...` `        ``if` `(n % 3 == 1) {` `            ``num_threes -= 1;` `            ``num_twos = 2;` `        ``}` `        ``// specifically for handling case where n = 5, 8,` `        ``// 11,` `        ``// ...` `        ``else` `if` `(n % 3 == 2) {` `            ``num_twos = 1;` `        ``}`   `        ``int` `res = 1;`   `        ``res *= (``int``)Math.Pow(3, num_threes);`   `        ``res *= (``int``)Math.Pow(2, num_twos);`   `        ``return` `res;` `    ``}` `}`

## Javascript

 `function` `maxProd(n) {` `    ``// edge case` `    ``if` `((n == 2) || (n == 3)) {` `        ``return` `n - 1;` `    ``}`   `    ``// how many times we can subtract 3 from n is (n/3)` `    ``let num_threes = Math.floor(n / 3);`   `    ``let num_twos = 0;`   `    ``// specifically for handling case where n = 4, 7, 10,` `    ``// ...` `    ``if` `(n % 3 == 1) {` `        ``num_threes -= 1;` `        ``num_twos = 2;` `    ``}` `    ``// specifically for handling case where n = 5, 8, 11,` `    ``// ...` `    ``else` `if` `(n % 3 == 2) {` `        ``num_twos = 1;` `    ``}`   `    ``let res = 1;`   `    ``res *= Math.pow(3, num_threes);`   `    ``res *= Math.pow(2, num_twos);`   `    ``return` `res;` `}`   `console.log(``"Maximum Product is "` `+ maxProd(10));`

Output

`Maximum Product is 36`

Time Complexity: O(log(n)), the pow() function takes log(n) time complexity so overall time complexity is O(log(n)).
Auxiliary Space: O(1)

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