Merge pull request #162 from SamTheKorean/main

[SAM] Week 10 solutions

Author: SamTheKorean

graph-valid-tree/samthekorean.py

@@ -0,0 +1,25 @@
+# O(n+e) where n is number of nodes and e is the number of edges.
+# O(n+e) where n is number of nodes and e is the number of edges.
+class Solution:
+    def validTree(self, numNodes: int, connections: List[List[int]]) -> bool:
+        adjacencyList = [[] for _ in range(numNodes)]
+        for src, dst in connections:
+            adjacencyList[src].append(dst)
+            adjacencyList[dst].append(src)
+
+        visitedNodes = set()
+
+        def detectCycle(currentNode, previousNode):
+            if currentNode in visitedNodes:
+                return True
+            visitedNodes.add(currentNode)
+            for neighbor in adjacencyList[currentNode]:
+                if neighbor == previousNode:
+                    continue
+                if detectCycle(neighbor, currentNode):
+                    return True
+            return False
+
+        if detectCycle(0, -1):
+            return False
+        return len(visitedNodes) == numNodes

house-robber-ii/samthekorean.py

@@ -0,0 +1,30 @@
+# TC : O(n), where n is the number of houses.
+# SC : O(1)
+class Solution:
+    def rob(self, nums):
+        n = len(nums)
+        if n == 1:
+            return nums[0]
+        if n == 2:
+            return max(nums[0], nums[1])
+        if n == 3:
+            return max(nums[0], max(nums[1], nums[2]))
+
+        a = nums[0]
+        b = max(nums[0], nums[1])
+        c = -1
+        a1 = nums[1]
+        b1 = max(nums[1], nums[2])
+        c1 = -1
+
+        for i in range(2, n):
+            if i < n - 1:
+                c = max(nums[i] + a, b)
+                a = b
+                b = c
+            if i > 2:
+                c1 = max(nums[i] + a1, b1)
+                a1 = b1
+                b1 = c1
+
+        return max(c, c1)

house-robber/samthekorean.py

@@ -0,0 +1,19 @@
+# TC : O(n)
+# SC : O(n)
+class Solution:
+    def rob(self, nums):
+        n = len(nums)
+        if n == 0:
+            return 0
+        if n == 1:
+            return nums[0]
+
+        dp = [-1] * n
+        dp[0] = nums[0]
+        if n > 1:
+            dp[1] = max(nums[0], nums[1])
+
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], nums[i] + dp[i - 2])
+
+        return dp[-1]

longest-palindromic-substring/samthekorean.py

@@ -0,0 +1,25 @@
+# O(n^2)
+# O(1)
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        if len(s) <= 1:
+            return s
+
+        def expand_from_center(left, right):
+            while left >= 0 and right < len(s) and s[left] == s[right]:
+                left -= 1
+                right += 1
+            return s[left + 1 : right]
+
+        max_str = s[0]
+
+        for i in range(len(s) - 1):
+            odd = expand_from_center(i, i)
+            even = expand_from_center(i, i + 1)
+
+            if len(odd) > len(max_str):
+                max_str = odd
+            if len(even) > len(max_str):
+                max_str = even
+
+        return max_str

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

@@ -0,0 +1,23 @@
+# O(n+e) where n is number of nodes and e is the number of edges.
+# O(n+e) where n is number of nodes and e is the number of edges.
+class Solution:
+    def countComponents(self, numNodes: int, connections: List[List[int]]) -> int:
+        adjacencyList = [[] for _ in range(numNodes)]
+        for src, dst in connections:
+            adjacencyList[src].append(dst)
+            adjacencyList[dst].append(src)
+
+        visitedNodes = set()
+
+        def depthFirstSearch(node):
+            visitedNodes.add(node)
+            for neighbor in adjacencyList[node]:
+                if neighbor not in visitedNodes:
+                    depthFirstSearch(neighbor)
+
+        componentCount = 0
+        for node in range(numNodes):
+            if node not in visitedNodes:
+                componentCount += 1
+                depthFirstSearch(node)
+        return componentCount