Merge pull request #2311 from ppxyn1/main

[ppxyn1] WEEK 13 solutions

Author: ppxyn1

insert-interval/ppxyn1.py

@@ -0,0 +1,32 @@
+# idea: - 
+# Time Complexity: O(n)
+class Solution:
+    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
+        res = []
+
+        # Ex,. newInterval = [2,5]
+        for i in range(len(intervals)):
+            start, end = intervals[i] 
+
+            # [2,5] > [1,5] 
+
+            # already passed
+            if end < newInterval[0]: 
+                res.append(intervals[i])
+
+            # not started yet
+            # (2)
+            elif newInterval[1] < start: 
+                res.append(newInterval) # [1,5]
+                for j in range(i, len(intervals)):
+                    res.append(intervals[j])
+                return res
+            else:
+                # (1)
+                # overlapping
+                newInterval[0] = min(newInterval[0], start)
+                newInterval[1] = max(newInterval[1], end)
+
+        res.append(newInterval)
+        return res
+

kth-smallest-element-in-a-bst/ppxyn1.py

@@ -0,0 +1,30 @@
+# 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 (inorder)
+#Time Complexity: O(n)
+
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        stack = []
+        cnt = 0
+        curr = root 
+        if not curr:
+            return
+
+        while curr or stack:
+            while curr:
+                stack.append(curr)
+                curr = curr.left 
+            curr = stack.pop()
+            cnt += 1
+
+            if cnt == k:
+                return curr.val
+            
+            curr = curr.right 
+

lowest-common-ancestor-of-a-binary-search-tree/ppxyn1.py

@@ -0,0 +1,23 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+# idea: BST property
+# Time Complexity : O(long n)
+class Solution:
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        cur = root
+        while cur:
+            if p.val > cur.val and q.val > cur.val:
+                cur = cur.right
+            elif p.val < cur.val and q.val < cur.val:
+                cur = cur.left
+            else:
+                return cur
+            
+
+
+