Merge pull request #2303 from doh6077/main

[doh6077] WEEK 12 solutions

Author: TonyKim9401

non-overlapping-intervals/doh6077.py

@@ -0,0 +1,47 @@
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+
+        # # first structure it as graph (dictionary)
+        # intervals = [[1,2],[2,3],[3,4],[1,3]]
+        # dict = {}
+        # for i in intervals:
+        #     dict[i[0]] = []
+        # for a,b in intervals:
+        #     dict[a].append(b)
+
+        # #Overlapping edge cases
+        # # 1. [1,4], [2,3] -> not sure how to figure out 
+        # # 2. [1,3], [1,2] , [3,4] -> check if the value has connection 
+        # # 3. [1,2], [1,2], [1,2] ( same values) use hashset 
+
+        # count = 0 
+        # visited = set()
+        # for key, value in dict.items():
+        #     if value not in visited:
+        #         visited.append(value)
+        #     else:
+        #         count += 1
+        # return count
+
+        # count = 0 
+        # visited = set()
+        # for a, b in intervals:
+        #     if b not in visited:
+        #         visited.add(b)
+        #     else:
+        #         count += 1
+        # return count 
+
+
+        # greedy Approach 
+        intervals.sort()
+
+        res = 0 
+        prevEnd = intervals[0][1]
+        for start, end in intervals[1:]:
+            if start >= prevEnd:
+                prevEnd = end
+            else:
+                res += 1
+                prevEnd = min(prevEnd, end)
+        return res

remove-nth-node-from-end-of-list/doh6077.py

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+class Solution:
+    # 19. Remove Nth Node From End of List
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        dummy = ListNode(0, head)
+        
+        #  Calculate Length 
+        length = 0
+        current = head
+        while current:
+            length += 1
+            current = current.next
+            
+        stop_index = length - n
+        
+        current = dummy 
+        for _ in range(stop_index):
+            current = current.next
+        
+        # Delete the node
+        current.next = current.next.next
+        
+        # Return the start of the list (dummy.next handles if head changed)
+        return dummy.next

same-tree/doh6077.py

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+# 
+# 100. Same Tree
+# https://leetcode.com/problems/same-tree/description/
+
+
+# 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
+class Solution:
+    #Time Complexity: O(N)- it visits every node once
+    #Space Complexity: O(H)- H is the height of the tree, which is the maximum depth of the recursion stack
+    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        # DFS approach
+        def dfs(node1, node2):
+            if node1 is None and node2 is None:
+                return True
+            if node1 is None or node2 is None:
+                return False 
+            if node1.val != node2.val:
+                return False 
+            return dfs(node1.left,node2.left) and dfs(node1.right,node2.right)
+
+        return dfs(p, q)