Program to find maximum profit after cutting rods and selling same length rods in Python

PythonServer Side ProgrammingProgramming

Suppose we have a list of rod lengths called rodLen. We also have another two integers called profit and cost, represents profit per length and cost per cut. We can make gain profit per unit length of a rod but we can only sell rods that are all of the same length. We can also cut a rod into two pieces such that their lengths are integers, but we have to pay cost amount for each cut. We can cut a rod as many times as we want. We have to find the maximum profit that we can make.

So, if the input is like rodLen = [7, 10] profit = 6 cost = 4, then the output will be 82, as we can cut the rod of length 7 into two rods, with length 5 and 2. We can then cut the rod of length 10 into two rods, both with length 5. Then sell all 3 rods of length 5 for a total profit of (5 + 5 + 5) * 6 - (2*4) = 82.

To solve this, we will follow these steps −

  • n := size of rodLen
  • if n is same as 0, then
    • return 0
  • l_max := maximum of rodLen
  • p_max := 0
  • for cuts in range 1 to l_max, do
    • p_cut := 0
    • for each rod_len in rodLen, do
      • if rod_len < cuts, then
        • go for next iteration
      • c_count := rod_len / cuts
      • total_len := c_count * cuts
      • if rod_len is same as total_len, then
        • c_count := c_count - 1
      • curr_profit := total_len * profit - cost * c_count
      • if curr_profit < 0, then
        • go for next iteration
      • p_cut := p_cut + curr_profit
    • p_max := maximum of p_max and p_cut
  • return p_max

Example

Let us see the following implementation to get better understanding −

def solve(rodLen, profit, cost):
   n = len(rodLen)
   if n == 0:
      return 0
   l_max = max(rodLen)
   p_max = 0

   for cuts in range(1, l_max + 1):
      p_cut = 0
      for rod_len in rodLen:
         if rod_len < cuts:
            continue
         c_count = rod_len // cuts
         total_len = c_count * cuts
         if rod_len == total_len:
            c_count -= 1
         curr_profit = total_len * profit - cost * c_count
         if curr_profit < 0:
            continue
         p_cut += curr_profit
      p_max = max(p_max, p_cut)
   return p_max

rodLen = [7, 10]
profit = 6
cost = 4
print(solve(rodLen, profit, cost))

Input

[7, 10], 6, 4

Output

82
raja
Updated on 18-Oct-2021 12:11:39

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