Additional notes on weekly prem

Author: Zanger67

my-submissions/m261.md

@@ -1,28 +1,21 @@
-class Solution:
-def validTree(self, n: int, edges: List[List[int]]) -> bool:
-if len(edges) != n - 1 :
-return False
+Notes
 
-        # Create edge map
-        e = defaultdict(set)
-        for u, v in edges :
-            e[u].add(v)
-            e[v].add(u)
+-   For a tree to be valid, $|E|=|V|-1$ must be true where $E$ is the edge set and $V$ is the vertex set
+-   Check for cycle starting from a random node -- track visited nodes
+    -   If any cycle is found, we return False
+    -   If no cycle is found, we continue
+-   Check if all nodes were visited
+    -   If all nodes visited, we return True
+    -   Else, we return False
 
-        # Detect cycle to save runtime -- breaks out early if cycle detected otherwise attempts
-        # to validate by visiting all n edges
-        def dfs_detect_cycle(curr: int, prev: int, e: defaultdict, visited: Set[int]) -> bool :
-            if curr in visited :
-                return True
-            visited.add(curr)
-            if any(dfs_detect_cycle(nxt, curr, e, visited) for nxt in e[curr] if nxt != prev) :
-                return (True, visited)
-            return False
+In theory the all nodes visited check is all that we need as long as avoid infinite looping in cycles. However, it's more efficient if we break out early if a cycle is found.
 
-        # Since dfs_detect_cycle() is called first before the or, visited() will be
-        # filled prior to len() being tested
-        visited = set()
-        if dfs_detect_cycle(1, None, e, visited) or not len(visited) == n :
-            return False
+The reason for the all nodes visited check after cycles is for graphs such as this:
 
-        return True
+```
+     1          2
+     |         / \
+     3        4---5
+```
+
+This graph has 5 vertices and 4 edges which fulfills the first condition. It's possible we'll start our cycle check on vertex 1 wherein we won't have a cycle, but there's a cycle elsewhere.