Colin toft

eating_cookies/eating_cookies.py

@@ -5,8 +5,23 @@
 # The cache parameter is here for if you want to implement
 # a solution that is more efficient than the naive 
 # recursive solution
+
+##eating cookies is going to allow us to figure out how many Ways there are to eating cookies 
+## n is effictbley cookies with 1,1,1 1,2 2,1 3
 def eating_cookies(n, cache=None):
-  pass
+  count = 0 
+  if n == 0 or n == 1:
+    return 1
+  elif n == 2:
+    return 2
+  else: 
+    arr = [1, 1, 2]
+    for number in range(3, n+1):
+      count = arr[-3] + arr[-2] + arr[-1]
+      arr.append(count)
+
+    return count
+
 
 if __name__ == "__main__":
   if len(sys.argv) > 1:

recipe_batches/recipe_batches.py

@@ -1,9 +1,23 @@
 #!/usr/bin/python
 
+## whats the question asking: how many batches of a recipe can we make from th input of 2 dictionaries dic1(recipes), dic2(ingreidents)
+## with this in mind we need to input both dictionaries and return the maxium batches possible may be //
+## https://www.w3schools.com/python/python_sets.asp
+##https://www.w3schools.com/python/ref_func_min.asp
+
 import math
 
 def recipe_batches(recipe, ingredients):
-  pass 
+    if set(recipe) == set(ingredients): ## check to see if we have the correct ammount of ingredients to make a single or multiple bacthes
+      batches =[]
+      for i in recipe:  ## check how many if any ammount of batches can be made from what we have avaible 
+          if ingredients[i] >= recipe[i]:
+            consumed = ingredients[i] // recipe[i]
+            batches.append(consumed)
+      return min(batches) ## this will show us what is the minium ammount of batches we can make 
+      ## we need to make a case senario for when there is not enough to make any batches using else and have the return be zero
+    else:
+      return 0 
 
 
 if __name__ == '__main__':

stock_prices/stock_prices.py

@@ -1,9 +1,24 @@
 #!/usr/bin/python
 
+### whats the problem we have a list of numbers, and we need to check which number is the biggest in the list then store and compare
+## this index to the number with the lowest index and then take the lowest from the highest
+
 import argparse
 
 def find_max_profit(prices):
-  pass
+    mp = [0] ## set max profit to initial 0 
+
+    for i in range(0, len(prices) -1): ## getting the initial value from mp to the last element
+      ci = i ## define index
+
+      for x in range(ci + 1, len(prices) -1): ## initiate the check from initial index to last index -1
+        if prices[x] - prices[ci] > mp[0]: ## subtract current index from the index we have in prices 
+          mp[0] = prices[x] - prices[ci]
+
+        elif prices[x] - prices[ci] < 0 and mp[0] == 0: ## if we hit negative numbers 
+          mp[0] = prices[x] - prices[ci]
+
+    return mp[0]
 
 
 if __name__ == '__main__':