Objective: Given a rope of length n meters, write an algorithm to cut the rope in such a way that the product of different lengths of rope is maximum. At least one cut has to be made.
Note: Assume that the length of the rope is more than 2 meters since at least one cut has to be made.
This is yet another problem where you will see the advantage of dynamic programming over recursion. I will show you how storing and reusing the results of sub-problems will reduce the complexity of an algorithm significantly. This kind of questions is asked in interviews like Amazon, Microsoft, Google, etc.
• Rope length: 2 • Options: (1*1) • Maximum Product Cutting : 1*1 = 1 • Rope length: 3 • Options: (1*1*1, 2*1) • Maximum Product Cutting : 2*1 = 2 • Rope length: 4 • Options: (1*1*1*1, 2*1*1, 3*1, 2*2) • Maximum Product Cutting : 2*2 = 4 • Rope length: 5 • Options: (1*1*1*1*1, 2*1*1*1, 3*1*1, 2*2*1, 3*2) • Maximum Product Cutting : 3*2 = 6
This problem is similar to the “Rod Cutting Problem“.
- Check the products of all possible cuts that can be made in the rope and return the maximum product.
- For every length, there are two options, either a cut to be made or not. Solve the problem for both options and choose the maximum.
- After Making the cut the further options are, Either this cut will produce the max product OR we need to make further cuts
Let MPC(n) is the maximum product for length n.
- MPC(n) = max(i*(n-i), i*MPC(n-i)) for all (i=1,2…..n) (After Making the cut the further options are , Either this cut will produce the max product OR we need to make further cuts).
Time Complexity: O(2n).
But yet again we are solving many sub-problems repeatedly.
We will solve this problem in a Bottom-Up manner.
We will store the solutions for sub-problems when it getting solved for the first time and use them again in the future so that we don’t have to solve them again.
Time Complexity: O(N^2)
See the Code for a better understanding.
Output: Dynamic Programming: Maximum product cutting in 10 is: 36