[ppxyn1] WEEK 10 solutions

invert-binary-tree/ppxyn1.py

@@ -0,0 +1,24 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+# idea : dfs
+# Time Complexity : O(n)
+
+class Solution:
+    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        if root is None:
+            return None
+
+        tmp = root.left
+        root.left = root.right 
+        root.right = tmp
+
+        self.invertTree(root.left)
+        self.invertTree(root.right)
+        return root
+
+

jump-game/ppxyn1.py

@@ -0,0 +1,22 @@
+# idea: recursive
+# Time Complexity : O(n^n) w.o cache
+#                 : O(n^2)   w cache
+
+class Solution:
+    def canJump(self, nums: List[int]) -> bool:
+        last_idx = len(nums) -1
+
+        @cache
+        def jump(i):
+            if i == last_idx:
+                return True
+            # jump from 1 to numbers                 
+            for j in range(1, nums[i] + 1):
+                if jump(i + j):
+                    return True
+
+            return False
+        return jump(0)
+
+
+

search-in-rotated-sorted-array/ppxyn1.py

@@ -0,0 +1,30 @@
+# idea : binary search
+# Time Complexity : O(log n)
+
+# Sorting is not allowed because it costs O(n log n), so we must handle it using binary search without sorting.
+# One way is to check where the sorted order is disrupted.
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        left, right = 0, len(nums)-1
+
+        while left <= right:
+            mid = (left + right) // 2
+
+            if nums[mid] == target:
+                return mid
+
+            if nums[left] <= nums[mid]:
+                if nums[left] <= target < nums[mid]:
+                    right = mid - 1
+                else:
+                    left = mid + 1
+
+            else:
+                if nums[mid] < target <= nums[right]:
+                    left = mid + 1
+                else:
+                    right = mid - 1
+        return -1
+
+