1971. Find if Path Exists in Graph - Leetcode Solution

Leetcode Problem Link Code Solution Link:

Common Setup

Base Case: If source == destination, return True, as no path is needed when starting at the destination.
Graph Construction: Create a defaultdict(list) called graph. For each edge [u, v] in edges, append v to graph[u] and u to graph[v], forming an undirected graph with bidirectional connections.
Visited Set: Initialize a set seen and add the source node to mark it as visited, preventing cycles during traversal.

💡Optimal Solution: Recursive DFS

This solution uses a recursive depth-first search to explore the graph, diving deep into each branch before backtracking.

DFS Function: Define a recursive function dfs(i) that takes a node i:
- If i == destination, return True, indicating the target is reached.
- For each neighbor nei_node in graph[i], if not in seen:
  - Add nei_node to seen.
  - Recursively call dfs(nei_node). If it returns True, propagate True.
- If no path is found, return False.
Execution: Call dfs(source) and return its result.
Why It Works: Recursive DFS explores each branch deeply, backtracking when necessary. The seen set ensures nodes are visited once, preventing infinite loops in the undirected graph. It’s concise but may face recursion depth issues for very deep graphs.

Code Solution (Recursive DFS)

# Recursive DFS
class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        if source == destination:
            return True
        
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        seen = set()
        seen.add(source)

        def dfs(i):
            if i == destination:
                return True
            
            for nei_node in graph[i]:
                if nei_node not in seen:
                    seen.add(nei_node)
                    if dfs(nei_node):
                        return True
            
            return False
        
        return dfs(source)

#include <vector>
#include <unordered_map>
#include <unordered_set>

using namespace std;

class Solution {
public:
    bool validPath(int n, vector<vector<int>>& edges, int source, int destination) {
        if (source == destination) return true;

        unordered_map<int, vector<int>> graph;
        for (const auto& edge : edges) {
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
        }

        unordered_set<int> seen;
        seen.insert(source);

        return dfs(source, destination, graph, seen);
    }

private:
    bool dfs(int node, int destination, unordered_map<int, vector<int>>& graph, unordered_set<int>& seen) {
        if (node == destination) return true;

        for (int neighbor : graph[node]) {
            if (seen.find(neighbor) == seen.end()) {
                seen.insert(neighbor);
                if (dfs(neighbor, destination, graph, seen)) return true;
            }
        }
        return false;
    }
};

import java.util.*;

public class Solution {
    public boolean validPath(int n, int[][] edges, int source, int destination) {
        if (source == destination) return true;

        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int[] edge : edges) {
            graph.computeIfAbsent(edge[0], k -> new ArrayList<>()).add(edge[1]);
            graph.computeIfAbsent(edge[1], k -> new ArrayList<>()).add(edge[0]);
        }

        Set<Integer> seen = new HashSet<>();
        seen.add(source);

        return dfs(source, destination, graph, seen);
    }

    private boolean dfs(int node, int destination, Map<Integer, List<Integer>> graph, Set<Integer> seen) {
        if (node == destination) return true;

        for (int neighbor : graph.getOrDefault(node, Collections.emptyList())) {
            if (!seen.contains(neighbor)) {
                seen.add(neighbor);
                if (dfs(neighbor, destination, graph, seen)) return true;
            }
        }
        return false;
    }
}

// Recursive DFS
var validPath = function(n, edges, source, destination) {
    if (source === destination) return true;

    const graph = new Map();
    for (const [u, v] of edges) {
        if (!graph.has(u)) graph.set(u, []);
        if (!graph.has(v)) graph.set(v, []);
        graph.get(u).push(v);
        graph.get(v).push(u);
    }

    const seen = new Set();
    seen.add(source);

    const dfs = (node) => {
        if (node === destination) return true;

        for (const neighbor of graph.get(node) || []) {
            if (!seen.has(neighbor)) {
                seen.add(neighbor);
                if (dfs(neighbor)) return true;
            }
        }
        return false;
    };

    return dfs(source);
}

💡Optimal Solution 2: Iterative DFS with Stack

This solution uses an iterative depth-first search with a stack, mimicking the recursive approach without recursion overhead.

Stack Initialization: Initialize a stack with the source node.
Iterative Traversal: While the stack is not empty:
- Pop a node (node) from the stack.
- If node == destination, return True.
- For each neighbor nei_node in graph[node], if not in seen:
  - Add nei_node to seen.
  - Push nei_node onto the stack.
Return: If the stack empties, return False.
Why It Works: The stack maintains the DFS exploration order, pushing unvisited neighbors to explore later, similar to recursive DFS. It avoids recursion stack limitations, making it robust for deep graphs, while the seen set prevents cycles.

Code Solution (Iterative DFS with Stack)

class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        if source == destination:
            return True
        
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        seen = set()
        seen.add(source)
        stack = [source]

        while stack:
            node = stack.pop()
            if node == destination:
                return True
            for nei_node in graph[node]:
                if nei_node not in seen:
                    seen.add(nei_node)
                    stack.append(nei_node)
        
        return False

#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <stack>

using namespace std;

class Solution {
public:
    bool validPath(int n, vector<vector<int>>& edges, int source, int destination) {
        if (source == destination) return true;

        unordered_map<int, vector<int>> graph;
        for (const auto& edge : edges) {
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
        }

        unordered_set<int> seen;
        stack<int> stack;
        stack.push(source);
        seen.insert(source);

        while (!stack.empty()) {
            int node = stack.top();
            stack.pop();
            if (node == destination) return true;

            for (int neighbor : graph[node]) {
                if (seen.find(neighbor) == seen.end()) {
                    seen.insert(neighbor);
                    stack.push(neighbor);
                }
            }
        }
        return false;
    }
};

import java.util.*;

public class Solution {
    public boolean validPath(int n, int[][] edges, int source, int destination) {
        if (source == destination) return true;

        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int[] edge : edges) {
            graph.computeIfAbsent(edge[0], k -> new ArrayList<>()).add(edge[1]);
            graph.computeIfAbsent(edge[1], k -> new ArrayList<>()).add(edge[0]);
        }

        Set<Integer> seen = new HashSet<>();
        Stack<Integer> stack = new Stack<>();
        stack.push(source);
        seen.add(source);

        while (!stack.isEmpty()) {
            int node = stack.pop();
            if (node == destination) return true;
            for (int neighbor : graph.getOrDefault(node, Collections.emptyList())) {
                if (!seen.contains(neighbor)) {
                    seen.add(neighbor);
                    stack.push(neighbor);
                }
            }
        }
        return false;
    }
}

var validPath = function(n, edges, source, destination) {
    if (source === destination) return true;

    const graph = new Map();
    for (const [u, v] of edges) {
        if (!graph.has(u)) graph.set(u, []);
        if (!graph.has(v)) graph.set(v, []);
        graph.get(u).push(v);
        graph.get(v).push(u);
    }

    const seen = new Set();
    const stack = [source];
    seen.add(source);

    while (stack.length > 0) {
        const node = stack.pop();
        if (node === destination) return true;

        for (const neighbor of graph.get(node) || []) {
            if (!seen.has(neighbor)) {
                seen.add(neighbor);
                stack.push(neighbor);
            }
        }
    }

    return false;
};

💡Optimal Solution 3: BFS with Queue

This solution uses a breadth-first search with a queue, exploring nodes level by level from the source.

Queue Initialization: Initialize a deque (q) and append the source node.
BFS Traversal: While q is not empty:
- Remove the front node (node) using popleft().
- If node == destination, return True.
- For each neighbor nei_node in graph[node], if not in seen:
  - Add nei_node to seen.
  - Append nei_node to the queue.
Return: If the queue empties, return False.
Why It Works: BFS explores nodes in order of increasing distance from the source, ensuring efficient path discovery. The seen set prevents cycles, and the queue manages level-by-level traversal, making it ideal for finding paths (potentially the shortest in terms of edges).

Code Solution (BFS with Queue)

from collections import deque
class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        if source == destination:
            return True
        
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        seen = set()
        seen.add(source)
        q = deque()
        q.append(source)

        while q:
            node = q.popleft()
            if node == destination:
                return True
            for nei_node in graph[node]:
                if nei_node not in seen:
                    seen.add(nei_node)
                    q.append(nei_node)
        
        return False # Time: O(N + E), Space: O(N + E)

#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <queue>

using namespace std;

class Solution {
public:
    bool validPath(int n, vector<vector<int>>& edges, int source, int destination) {
        if (source == destination) return true;

        unordered_map<int, vector<int>> graph;
        for (const auto& edge : edges) {
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
        }

        unordered_set<int> seen;
        queue<int> queue;
        queue.push(source);
        seen.insert(source);

        while (!queue.empty()) {
            int node = queue.front();
            queue.pop();
            if (node == destination) return true;

            for (int neighbor : graph[node]) {
                if (seen.find(neighbor) == seen.end()) {
                    seen.insert(neighbor);
                    queue.push(neighbor);
                }
            }
        }
        return false;
    }
};

import java.util.*;

public class Solution {
    public boolean validPath(int n, int[][] edges, int source, int destination) {
        if (source == destination) return true;

        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int[] edge : edges) {
            graph.computeIfAbsent(edge[0], k -> new ArrayList<>()).add(edge[1]);
            graph.computeIfAbsent(edge[1], k -> new ArrayList<>()).add(edge[0]);
        }

        Set<Integer> seen = new HashSet<>();
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(source);
        seen.add(source);

        while (!queue.isEmpty()) {
            int node = queue.poll();
            if (node == destination) return true;
            for (int neighbor : graph.getOrDefault(node, Collections.emptyList())) {
                if (!seen.contains(neighbor)) {
                    seen.add(neighbor);
                    queue.offer(neighbor);
                }
            }
        }
        return false;
    }
}

var validPath = function(n, edges, source, destination) {
    if (source === destination) return true;

    const graph = new Map();
    for (const [u, v] of edges) {
        if (!graph.has(u)) graph.set(u, []);
        if (!graph.has(v)) graph.set(v, []);
        graph.get(u).push(v);
        graph.get(v).push(u);
    }

    const seen = new Set();
    const queue = [source];
    seen.add(source);

    while (queue.length > 0) {
        const node = queue.shift();
        if (node === destination) return true;

        for (const neighbor of graph.get(node) || []) {
            if (!seen.has(neighbor)) {
                seen.add(neighbor);
                queue.push(neighbor);
            }
        }
    }

    return false;
};

Detailed Explanation

Problem Overview

The task is to determine whether there is a path between two nodes (source and destination) in an undirected graph. This is a fundamental graph traversal problem that can be solved using depth-first search (DFS), breadth-first search (BFS), or Union-Find. Since the graph is undirected, an edge [u, v] means both u is connected to v and v is connected to u.

Approach: Graph Traversal

The core idea is to traverse the graph starting from the source node and see if we can reach the destination. To do this, we first represent the graph using an adjacency list. For each edge [u, v], we record both u → v and v → u in a dictionary to capture the undirected nature of the graph.

After building the graph, we use a standard DFS approach. A stack is initialized with the source node and a set is used to track visited nodes to prevent cycles or repeated work. At each step, we pop a node from the stack and check if it matches the destination. If it does, we return true. If not, we push all unvisited neighbors of the current node onto the stack and continue. If the traversal ends without finding the destination, we return false.

Edge Cases

- If the source is equal to the destination, we return true immediately, as the path trivially exists.
- If there are no edges or the source node has no connections, we cannot proceed, and false is returned.

Time and Space Complexity

Time Complexity: O(V + E), where V is the number of nodes and E is the number of edges. We potentially visit every node and edge once.

Space Complexity: O(V + E) for storing the graph and visited set, and O(V) for the recursion stack or explicit stack used in DFS.

Conclusion

This problem is a classic example of graph traversal. DFS is a natural choice when we want to explore paths from one node to another, especially when we do not need the shortest path. The key to an efficient solution is using an adjacency list and tracking visited nodes to avoid unnecessary work.

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