Update rod_cutting.py (#995)

* Update rod_cutting.py

A hopefully clearer implementation without dependence on global variables.

* Update rod_cutting.py

added doctests

* Update rod_cutting.py

* Update rod_cutting.py

dynamic_programming/rod_cutting.py

@@ -1,58 +1,57 @@
-### PROBLEM ###
-"""
-We are given a rod of length n and we are given the array of prices, also of
-length n. This array contains the price for selling a rod at a certain length.
-For example, prices[5] shows the price we can sell a rod of length 5.
-Generalising, prices[x] shows the price a rod of length x can be sold.
-We are tasked to find the optimal solution to sell the given rod.
-"""
+from typing import List
 
-### SOLUTION ###
-"""
-Profit(n) = max(1<i<n){Price(n),Price(i)+Profit(n-i)}
+def rod_cutting(prices: List[int],length: int) -> int:
+	"""
+	Given a rod of length n and array of prices that indicate price at each length.
+	Determine the maximum value obtainable by cutting up the rod and selling the pieces
+	
+	>>> rod_cutting([1,5,8,9],4) 
+	10
+	>>> rod_cutting([1,1,1],3) 
+	3
+	>>> rod_cutting([1,2,3], -1)
+	Traceback (most recent call last):
+   	ValueError: Given integer must be greater than 1, not -1
+   	>>> rod_cutting([1,2,3], 3.2)
+	Traceback (most recent call last):
+   	TypeError: Must be int, not float
+   	>>> rod_cutting([], 3)
+	Traceback (most recent call last):
+   	AssertionError: prices list is shorted than length: 3
+	
+	
 
-When we receive a rod, we have two options:
-a) Don't cut it and sell it as is (receiving prices[length])
-b) Cut it and sell it in two parts. The length we cut it and the rod we are
-left with, which we have to try and sell separately in an efficient way.
-Choose the maximum price we can get.
-"""
+	Args:
+		prices: list indicating price at each length, where prices[0] = 0 indicating rod of zero length has no value
+	    length: length of rod
 
-def CutRod(n):
-    if(n == 1):
-        #Cannot cut rod any further
-        return prices[1]
+	Returns:
+	    Maximum revenue attainable by cutting up the rod in any way. 
+	"""
 
-    noCut = prices[n] #The price you get when you don't cut the rod
-    yesCut = [-1 for x in range(n)] #The prices for the different cutting options
+	prices.insert(0, 0)
+	if not isinstance(length, int):
+	    raise TypeError('Must be int, not {0}'.format(type(length).__name__))
+	if length < 0: 
+	    raise ValueError('Given integer must be greater than 1, not {0}'.format(length))
+	assert len(prices) - 1 >= length, "prices list is shorted than length: {0}".format(length) 
 
-    for i in range(1,n):
-        if(solutions[i] == -1):
-            #We haven't calulated solution for length i yet.
-            #We know we sell the part of length i so we get prices[i].
-            #We just need to know how to sell rod of length n-i
-            yesCut[i] = prices[i] + CutRod(n-i)
-        else:
-            #We have calculated solution for length i.
-            #We add the two prices.
-            yesCut[i] = prices[i] + solutions[n-i]
+	return rod_cutting_recursive(prices, length)
 
-    #We need to find the highest price in order to sell more efficiently.
-    #We have to choose between noCut and the prices in yesCut.
-    m = noCut #Initialize max to noCut
-    for i in range(n):
-        if(yesCut[i] > m):
-            m = yesCut[i]
-
-    solutions[n] = m
-    return m
+def rod_cutting_recursive(prices: List[int],length: int) -> int:
+	#base case 
+	if length == 0:
+		return 0
+	value = float('-inf')
+	for firstCutLocation in range(1,length+1):
+		value = max(value, prices[firstCutLocation]+rod_cutting_recursive(prices,length - firstCutLocation))
+	return value
 
 
+def main():
+	assert rod_cutting([1,5,8,9,10,17,17,20,24,30],10) == 30
+	# print(rod_cutting([],0))
 
-### EXAMPLE ###
-length = 5
-#The first price, 0, is for when we have no rod.
-prices = [0, 1, 3, 7, 9, 11, 13, 17, 21, 21, 30]
-solutions = [-1 for x in range(length+1)]
+if __name__ == '__main__':
+	main()
 
-print(CutRod(length))