[ppxyn1] WEEK 02 solutions

climbing-stairs/ppxyn1.py

@@ -1,6 +1,33 @@
 # idea: DP
-# I'm not always sure how to approach DP problems. I just try working through a few examples step by step and then check that it would be DP.
-# If you have any suggestions for how I can come up with DP, I would appreciate your comments :)
+# Time Complexity : O(n)
+'''
+n = 4
+-------------------------------
+(dp[3] + 1step)
+1step + 1step + 1step + 1step 
+1step + 2step + 1step
+2step + 1step + 1step
+
+(dp[2] + 2step)
+1step + 1step + 2step
+2step + 2step
+
+================================
+n = 5
+--------------------------------
+(dp[4] + 1step)
+1step + 1step + 1step + 1step + 1step
+1step + 2step + 1step + 1step
+1step + 1step + 2step + 1step
+2step + 1step + 1step + 1step
+2step + 2step + 1step
+
+(dp[3] + 2step)
+1step + 2step + 2step
+2step + 1step + 2step
+1step + 1step + 1step + 2step
+'''
+
 
 class Solution:
     def climbStairs(self, n: int) -> int:
@@ -9,7 +36,6 @@ def climbStairs(self, n: int) -> int:
         dp = [0] * (n+1)
         dp[2], dp[3] = 2, 3
 
-        #for i in range(4, n): error when n=4       
         for i in range(4, n+1):
             dp[i] = dp[i-1] + dp[i-2]
 

valid-anagram/ppxyn1.py

@@ -1,27 +1,26 @@
-#idea: dictionary
 from collections import Counter 
+# class Solution:
+#     def isAnagram(self, s: str, t: str) -> bool:
+#         s_dic = Counter(sorted(s))
+#         t_dic = Counter(sorted(t))
+#         return s_dic==t_dic
 
-class Solution:
-    def isAnagram(self, s: str, t: str) -> bool:
-        s_dic = Counter(sorted(s))
-        t_dic = Counter(sorted(t))
-        print(s_dic, t_dic)
-        return s_dic==t_dic
+# fixed
+# do not need to both Counter and sorting 
 
-
-
-# Trial and Error
-'''
-When you call sorted() on a dictionary, it only extracts and sorts the keys,and the values are completely ignored.
+# Ans 1 (Only Counter)
+# Time Complexity: O(N)
 class Solution:
     def isAnagram(self, s: str, t: str) -> bool:
         s_dic = Counter(s)
         t_dic = Counter(t)
-        print(s_dic, t_dic)
-        return sorted(s_dic)==sorted(t_dic)
-'''
-
-
-
+        return s_dic==t_dic
 
+# Ans 2  (Only Sorting)
+# Time Complexity: O(Nlog(N))
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        s_arr = sorted(s)
+        t_arr = sorted(t)
+        return s_arr==t_arr