CSPT13-Karthik-SC-Algorithms

Short-Answer/Algorithms_Answers.md

@@ -2,14 +2,27 @@
 
 ## Exercise I
 
-a)
+a) O(n) -- run time increases linearly as n increases. Single loop.
 
 
-b)
+b) O(n^2) -- 2 loops. So run time increases twice as n increases. 
 
 
-c)
+c)  It is O(n) because the runtime will increase at a constant pace with the size of the input. 
 
 ## Exercise II
 
+Binary search is best way to go about this. 
+
+1. Drop the egg on floor 1, if it breaks then f =0 and end program
+
+2. if egg does not break. 
+    halve the total num of floors, if total floors is odd then move to 1 floor lower. 
+        drop egg
+            if it breaks, consider this floor as top floor. Move to middle of 1st floor and current floor. 
+            if its does not break, consider this as bottom floor. move to middle floor of topmost and current floor. 
+
+3. follow above process and find the floor where egg does not break, move 1 floor ahead and find where it does not. 
+
+4. this is give the 'f' answer. 
 

recursive_count_th/count_th.py

@@ -4,7 +4,10 @@
 Your function must utilize recursion. It cannot contain any loops.
 '''
 def count_th(word):
+    if len(word)<2: # if word length is less than 2 return 0 as we are looking for 'th'
+        return 0
+    if word[0:2] == 'th':              # use recursion to repeated checks for 'th'
+        return count_th(word[1:]) + 1
 
-    # TBC
-
-    pass
+    return count_th(word[1:])
+

robot_sort/robot_sort.py

@@ -96,14 +96,50 @@ def sort(self):
         """
         Sort the robot's list.
         """
-        # Fill this out
-        pass
+        # in this algo I am using the light for exit loop, or know when to quit sorting. 
+        # motivation in this algo is to reduce the number of steps robot is taking to get the sorting done. 
+        # idea is not to reduce the code execution time. Also we want to keep space complexity as 1, since making more 
+           # copies is not an option. So recursive sorting is ruled out. I am doing an iterative approach. 
+
+        # Plan:
+          # 1. Robot picks up first ele, moves right if possible and compares. 
+          # 2. if comparioson result is 1, swap. move left by 1 step and swap again. set light ON
+          # 3. if comparision result is not 1, move left 1 step, but back the carrying element, move 1 step to right. 
+               # No action with light. 
+          # 4. repeat steps 1-3, until robot cannot move right. 
+          # 5. Move back to first element after 1 pass and repeat steps 1-4 
+          # 6. exit when after 1 pass through the array light remains off. 
+
+        self.set_light_on() # set light on in beginning. 
+
+        while self.light_is_on():
+            self.set_light_off()
+
+            while self.can_move_right(): # Step 1
+                self.swap_item()
+                self.move_right()
+
+                if self.compare_item() == 1: # Step 2 
+                    self.swap_item()
+                    self.move_left()
+                    self.swap_item()
+                    self.move_right()
+                    self.set_light_on()
+
+                else:                       # Step 3 
+                    self.move_left()
+                    self.swap_item()
+                    self.move_right()
+
+            while self.can_move_left():  # step 5 
+                self.move_left() 
 
 
 if __name__ == "__main__":
     # Test our your implementation from the command line
     # with `python robot_sort.py`
 
+
     l = [15, 41, 58, 49, 26, 4, 28, 8, 61, 60, 65, 21, 78, 14, 35, 90, 54, 5, 0, 87, 82, 96, 43, 92, 62, 97, 69, 94, 99, 93, 76, 47, 2, 88, 51, 40, 95, 6, 23, 81, 30, 19, 25, 91, 18, 68, 71, 9, 66, 1, 45, 33, 3, 72, 16, 85, 27, 59, 64, 39, 32, 24, 38, 84, 44, 80, 11, 73, 42, 20, 10, 29, 22, 98, 17, 48, 52, 67, 53, 74, 77, 37, 63, 31, 7, 75, 36, 89, 70, 34, 79, 83, 13, 57, 86, 12, 56, 50, 55, 46]
 
     robot = SortingRobot(l)