LC-1971 Easy LeetCode

1971. Find if Path Exists in Graph

Depth-First Search Breadth-First Search Union Find Graph

Read the full problem statement on LeetCode.

Difficulty: easy Acceptance: 54% Topics: Depth-First Search, Breadth-First Search, Union Find, Graph

Reading material

Trees & Traversals 18 min Union–Find (DSU) 12 min Graphs 16 min

Reference solution (spoiler · python)

# Time:  O(|V| + |E|)
# Space: O(|V| + |E|)

import collections


# bi-bfs solution
class Solution(object):
    def validPath(self, n, edges, start, end):
        """
        :type n: int
        :type edges: List[List[int]]
        :type start: int
        :type end: int
        :rtype: bool
        """
        def bi_bfs(adj, start, target):
            left, right = {start}, {target}
            lookup = set()
            steps = 0
            while left:
                for pos in left:
                    lookup.add(pos)
                new_left = set()
                for pos in left:
                    if pos in right: 
                        return steps
                    for nei in adj[pos]:
                        if nei in lookup:
                            continue
                        new_left.add(nei)
                left = new_left
                steps += 1
                if len(left) > len(right): 
                    left, right = right, left
            return -1

        adj = collections.defaultdict(list)
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)
        return bi_bfs(adj, start, end) >= 0


# Time:  O(|V| + |E|)
# Space: O(|V| + |E|)
# bfs solution
class Solution2(object):
    def validPath(self, n, edges, start, end):
        """
        :type n: int
        :type edges: List[List[int]]
        :type start: int
        :type end: int
        :rtype: bool
        """
        def bfs(adj, start, target):
            q = [start]
            lookup = set(q)
            steps = 0
            while q:
                new_q = []
                for pos in q:
                    if pos == target:
                        return steps
                    for nei in adj[pos]:
                        if nei in lookup:
                            continue
                        lookup.add(nei)
                        new_q.append(nei)
                q = new_q
                steps += 1
            return -1  

        adj = collections.defaultdict(list)
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)
        return bfs(adj, start, end) >= 0


# Time:  O(|V| + |E|)
# Space: O(|V| + |E|)
# dfs solution
class Solution3(object):
    def validPath(self, n, edges, start, end):
        """
        :type n: int
        :type edges: List[List[int]]
        :type start: int
        :type end: int
        :rtype: bool
        """
        def dfs(adj, start, target):
            stk = [start]
            lookup = set(stk)
            while stk:
                pos = stk.pop()
                if pos == target:
                    return True
                for nei in reversed(adj[pos]):
                    if nei in lookup:
                        continue
                    lookup.add(nei)
                    stk.append(nei)
            return False 

        adj = collections.defaultdict(list)
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)
        return dfs(adj, start, end)

Solution from kamyu104/LeetCode-Solutions · MIT