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+#!/usr/bin/python2
+
+# Fuel Injection Perfection
+# =========================
+
+# Commander Lambda has asked for your help to refine the automatic quantum
+# antimatter fuel injection system for her LAMBCHOP doomsday device. It's a
+# great chance for you to get a closer look at the LAMBCHOP - and maybe sneak
+# in a bit of sabotage while you're at it - so you took the job gladly.
+
+# Quantum antimatter fuel comes in small pellets, which is convenient since the
+# many moving parts of the LAMBCHOP each need to be fed fuel one pellet at a
+# time. However, minions dump pellets in bulk into the fuel intake. You need to
+# figure out the most efficient way to sort and shift the pellets down to a
+# single pellet at a time.
+
+# The fuel control mechanisms have three operations: 
+
+# 1) Add one fuel pellet
+
+# 2) Remove one fuel pellet
+
+# 3) Divide the entire group of fuel pellets by 2 (due to the destructive
+# energy released when a quantum antimatter pellet is cut in half, the safety
+# controls will only allow this to happen if there is an even number of
+# pellets)
+
+# Write a function called solution(n) which takes a positive integer as a
+# string and returns the minimum number of operations needed to transform the
+# number of pellets to 1. The fuel intake control panel can only display a
+# number up to 309 digits long, so there won't ever be more pellets than you
+# can express in that many digits.
+
+# For example:
+# solution(4) returns 2: 4 -> 2 -> 1
+# solution(15) returns 5: 15 -> 16 -> 8 -> 4 -> 2 -> 1
+
+def change(b, n):
+  a = int("".join([str(x) for x in b]), 2) + n
+  return [int(i) for i in list("{0:b}".format(a))]
+
+def solution(n):
+  if n == "1": return 0
+  if n == "2": return 1
+  
+  b = [int(i) for i in list("{0:b}".format(int(n, 10)))]
+
+  c = 0
+  
+  flag = True
+  while flag == True:
+    if (len(b) <= 2):
+      flag = False
+      continue
+    if b[-1] == 0:
+      c += 1
+      b.pop()
+    else:
+      if b[-2] == 1:
+        c += 1
+        b = change(b, 1)
+      else:
+        c += 1
+        b = change(b, -1)
+
+  if b[1] == 1:
+    c += 2
+  else:
+    c += 1
+
+  return c
+
+def funny_add1(b):
+  b.reverse()
+
+  carry = 0
+  for i in range(0, len(b)):
+    if b[i] == 0:
+      b[i] = 1
+      break
+    else: #if b[i] == 1:
+      b[i] = 0
+      carry = 1
+    if i == len(b) - 1 and carry == 1:
+      b.append(1)
+
+  b.reverse()
+  return b
+
+print(solution("19"))
+print(solution("4")) # => 2
+print(solution("15")) # => 5