[dolphinflow86] WEEK 13 Solutions (#2864)

* insert interval solution

* lowest common ancestor of a binary search tree solution

* kth smallest element in a bst solution

* find median from data stream solution

* serialize and deserialize binary tree solution

* number of connected components in an undirected graph solution

Author: dolphinflow86

find-median-from-data-stream/dolphinflow86.py

@@ -0,0 +1,32 @@
+# N is the number of elements added to the data stream.
+# TC: O(log N) for addNum, O(1) for findMedian
+# SC: O(N) - stores elements split into two heaps
+
+import heapq
+
+
+class MedianFinder:
+
+    def __init__(self):
+        self.small = []  # max-heap (store negative values)
+        self.large = []  # min-heap
+
+    def addNum(self, num: int) -> None:
+        heapq.heappush(self.small, -num)
+
+        if self.small and self.large and (-self.small[0] > self.large[0]):
+            val = -heapq.heappop(self.small)
+            heapq.heappush(self.large, val)
+
+        if len(self.small) > len(self.large) + 1:
+            val = -heapq.heappop(self.small)
+            heapq.heappush(self.large, val)
+
+        if len(self.large) > len(self.small):
+            val = heapq.heappop(self.large)
+            heapq.heappush(self.small, -val)
+
+    def findMedian(self) -> float:
+        if len(self.small) > len(self.large):
+            return float(-self.small[0])
+        return (-self.small[0] + self.large[0]) / 2.0

insert-interval/dolphinflow86.py

@@ -0,0 +1,28 @@
+# N is the number of intervals.
+# TC: O(N) - single pass through intervals array
+# SC: O(N) - stores result intervals list
+
+
+class Solution:
+
+    def insert(self, intervals, newInterval):
+        result = []
+        i = 0
+        n = len(intervals)
+
+        while i < n and intervals[i][1] < newInterval[0]:
+            result.append(intervals[i])
+            i += 1
+
+        while i < n and intervals[i][0] <= newInterval[1]:
+            newInterval[0] = min(newInterval[0], intervals[i][0])
+            newInterval[1] = max(newInterval[1], intervals[i][1])
+            i += 1
+
+        result.append(newInterval)
+
+        while i < n:
+            result.append(intervals[i])
+            i += 1
+
+        return result

kth-smallest-element-in-a-bst/dolphinflow86.py

@@ -0,0 +1,32 @@
+# H is the height of the BST, and K is the target rank.
+# TC: O(H + K) - in-order traversal stops after visiting K elements
+# SC: O(H) - stack memory for in-order traversal recursion/loop
+
+# 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:
+
+    def kthSmallest(self, root, k: int) -> int:
+        stack = []
+        curr = root
+
+        while curr or stack:
+            while curr:
+                stack.append(curr)
+                curr = curr.left
+
+            curr = stack.pop()
+            k -= 1
+
+            if k == 0:
+                return curr.val
+
+            curr = curr.right
+
+        return -1

lowest-common-ancestor-of-a-binary-search-tree/dolphinflow86.py

@@ -0,0 +1,26 @@
+# H is the height of the binary search tree.
+# TC: O(H) - traverses down tree height
+# SC: O(1) - uses iterative traversal without recursion stack
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+
+class Solution:
+
+    def lowestCommonAncestor(self, root, p, q):
+        curr = root
+
+        while curr:
+            if p.val < curr.val and q.val < curr.val:
+                curr = curr.left
+            elif p.val > curr.val and q.val > curr.val:
+                curr = curr.right
+            else:
+                return curr
+
+        return None

number-of-connected-components-in-an-undirected-graph/dolphinflow86.py

@@ -0,0 +1,34 @@
+# n is the number of nodes, and E is the number of edges.
+# TC: O(V + E * alpha(V)) - near linear time with path compression Union-Find
+# SC: O(V) - parent and rank arrays for Union-Find
+
+
+class Solution:
+
+    def countComponents(self, n: int, edges) -> int:
+        parent = list(range(n))
+        rank = [1] * n
+
+        def find(node):
+            if parent[node] != node:
+                parent[node] = find(parent[node])
+            return parent[node]
+
+        def union(n1, n2):
+            p1, p2 = find(n1), find(n2)
+            if p1 == p2:
+                return 0
+
+            if rank[p1] > rank[p2]:
+                parent[p2] = p1
+                rank[p1] += rank[p2]
+            else:
+                parent[p1] = p2
+                rank[p2] += rank[p1]
+            return 1
+
+        components = n
+        for u, v in edges:
+            components -= union(u, v)
+
+        return components

serialize-and-deserialize-binary-tree/dolphinflow86.py

@@ -0,0 +1,44 @@
+# N is the number of nodes in the binary tree.
+# TC: O(N) - visits each node once during serialization and deserialization
+# SC: O(N) - stores node values and recursion stack
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+
+class Codec:
+
+    def serialize(self, root):
+        vals = []
+
+        def dfs(node):
+            if not node:
+                vals.append("N")
+                return
+            vals.append(str(node.val))
+            dfs(node.left)
+            dfs(node.right)
+
+        dfs(root)
+        return ",".join(vals)
+
+    def deserialize(self, data):
+        vals = data.split(",")
+        self.i = 0
+
+        def dfs():
+            if vals[self.i] == "N":
+                self.i += 1
+                return None
+
+            node = TreeNode(int(vals[self.i]))
+            self.i += 1
+            node.left = dfs()
+            node.right = dfs()
+            return node
+
+        return dfs()