Runtime: 47 ms (Top 48.8%) | Memory: 16.22 MB (Top 87.1%)

Author: AnasImloul

scripts/algorithms/W/Wiggle Subsequence/Wiggle Subsequence.py

@@ -1,44 +1,24 @@
+# Runtime: 47 ms (Top 48.8%) | Memory: 16.22 MB (Top 87.1%)
+
+#####################################################################################################################
+# Problem:  Wiggle Subsequence
+# Solution : Dynamic Programming
+# Time Complexity : O(n) 
+# Space Complexity : O(1)
+#####################################################################################################################
+
 class Solution:
     def wiggleMaxLength(self, nums: List[int]) -> int:
-        # Top down with memo
-        # Time: O(N2)
-        dp = {}
-        def wiggle(prevIndex, curIndex, sign):
-            if curIndex >= len(nums):
-                return 0
-            
-            if (prevIndex, curIndex, sign) in dp:
-                return dp[prevIndex, curIndex, sign]
-            
-            diff = nums[prevIndex] - nums[curIndex]
-            cur_sign = (diff > 0)
-            include = exclude = 0
-            
-            # if sign is different, we have two option, either take that element or ignore it
-            if diff != 0 and (cur_sign is not sign):
-                include = 1 + wiggle(curIndex, curIndex+1, cur_sign)
-                exclude = wiggle(prevIndex, curIndex+1, sign)
-            # since sign is same, here we have only one option, to exclude th curIndex
-            else:
-                # we can merge both exclude together to reduce lines, but keeping it separate for readability
-                exclude = wiggle(prevIndex, curIndex+1, sign)
-            
-            dp[prevIndex, curIndex, sign] = max(include, exclude)
-            return dp[prevIndex, curIndex, sign]
-              
-        return wiggle(0, 1, None) + 1
 
-        #--------------------------------------------------------------------------#
+        positive, negative = 1, 1
+        
+        if len(nums) < 2:
+            return len(nums)
 
-        # Greedy Time: O(N)
-        # Count all the changes of direction (peaks and vallyes)
-        count = 1
-        sign = 0
         for i in range(1, len(nums)):
-            if nums[i-1] < nums[i] and sign != 1:
-                sign = 1
-                count += 1
-            elif nums[i-1] > nums[i] and sign != -1:
-                sign = -1
-                count +=1
-        return count
+            if nums[i] > nums[i - 1]:
+                positive = negative + 1
+            elif nums[i] < nums[i - 1]:
+                negative = positive + 1
+                
+        return max(positive, negative)