1697. Checking Existence of Edge Length Limited Paths: A Deep Dive

When tackling advanced graph problems in competitive programming, efficiency is paramount. One such challenge that frequently tests a developer's understanding of algorithms and data structures is LeetCode problem 1697. Checking Existence of Edge Length Limited Paths. Mastering this specific problem involves a deep dive into powerful techniques like offline processing and the Disjoint Set Union (DSU) data structure, offering a significant learning opportunity for anyone looking to enhance their algorithmic toolkit.

Understanding the Problem: 1697. Checking Existence of Edge Length Limited Paths

The problem asks us to consider an undirected graph with n nodes, numbered from 0 to n-1. We are given a list of edges, where each edge [u, v, weight] indicates a bidirectional connection between nodes u and v with a specified weight. Additionally, we receive a list of queries, where each query [p, q, limit] asks if it's possible to find a path between node p and node q such that every edge along that path has a weight strictly less than limit. Our goal is to return a boolean array answer, where answer[i] is true if a valid path exists for the i-th query, and false otherwise.

Let's break down the input and output requirements:

• n: An integer representing the number of nodes in the graph.
• edges: A 2D integer array where edges[i] = [u_i, v_i, w_i] denotes an undirected edge between u_i and v_i with weight w_i.
• queries: A 2D integer array where queries[j] = [p_j, q_j, limit_j] represents the j-th query.
• answer: A boolean array of the same length as queries, where answer[j] is the result for queries[j].

The core challenge lies in efficiently handling the limit constraint. A simple pathfinding algorithm like Breadth-First Search (BFS) or Depth-First Search (DFS) would work for a single query, but with potentially many queries and large graphs, this approach quickly becomes too slow. Other shortest path algorithms like Dijkstra's Algorithm or Floyd-Warshall are also essential in graph theory, but they typically find paths minimizing total weight or any path, not necessarily respecting a per-edge limit. We need a more sophisticated strategy that can leverage the structure of the problem.

Example Scenario

Imagine a graph with 3 nodes and edges [[0, 1, 2], [1, 2, 4]]. If we have a query [0, 2, 5]: Is there a path from 0 to 2 where all edges are < 5? Yes, 0 -> 1 (weight 2 < 5) and 1 -> 2 (weight 4 < 5). So, true.

If we have a query [0, 2, 3]: Is there a path from 0 to 2 where all edges are < 3? 0 -> 1 (weight 2 < 3) is fine. 1 -> 2 (weight 4) is NOT < 3. So, the path 0 -> 1 -> 2 is invalid under this limit. There's no other path. So, false.

This simple example highlights why the limit is critical and must be strictly adhered to for every edge in the path.

Prerequisites for Tackling Edge Length Limited Paths

Before diving into the optimal solution, it's essential to have a solid grasp of a few fundamental concepts in graph theory and data structures. These prerequisites will form the building blocks of our approach.

Graph Theory Basics

Understanding graphs, nodes, edges, weights, and connectivity is foundational. You should be familiar with what a path is and how to traverse a graph. For this problem, the graph is undirected, meaning edges can be traversed in both directions.

Sorting Algorithms

The efficient solution relies heavily on sorting both the edges of the graph and the queries themselves. Knowledge of comparison-based sorting algorithms like Merge Sort or Quick Sort, and understanding their O(N log N) time complexity, is crucial. We will sort both by edge weights and query limits.

Disjoint Set Union (DSU) Data Structure

The Disjoint Set Union (DSU), also known as Union-Find, is a data structure that keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets. It performs two primary operations:

• find(i): Determines which subset an element i is in. It returns a "representative" element of i's subset. This operation is often optimized with path compression, where all nodes on the path from i to its root are re-pointed directly to the root, flattening the tree and speeding up future find operations.
• union(i, j): Merges the subsets containing elements i and j into a single subset. This operation is often optimized with union by rank (or union by size), where the smaller tree is always attached to the root of the larger tree, preventing the DSU trees from becoming too tall and preserving the efficiency of find operations.

DSU is incredibly efficient for connectivity problems, especially when nodes are gradually connected. Optimized DSU implementations with path compression and union by rank/size achieve nearly constant time complexity (amortized O(alpha(N)), where alpha is the inverse Ackermann function, which is practically a very small constant, essentially < 5 for any realistic N) for find and union operations. This will be the cornerstone of our solution.

If you're new to DSU, it's highly recommended to review it before proceeding. Many online resources and textbooks offer excellent explanations and examples.

Inefficiencies of Naive Approaches

To appreciate the optimal solution, let's first consider why simpler approaches fall short.

Brute-Force BFS/DFS for Each Query

A straightforward approach for each query [p, q, limit] would be to:

1. Construct a subgraph containing only edges whose weights are strictly less than limit.
2. Perform a BFS or DFS starting from p in this subgraph to see if q is reachable.

Time Complexity Analysis:

• For each query, constructing the subgraph might involve iterating through all E edges.
• A BFS/DFS on a graph with V nodes and E' edges takes O(V + E') time. In the worst case, E' could be O(E).
• If there are Q queries, the total time complexity would be O(Q * (E + V + E)), which simplifies to O(Q * (V + E)).

Space Complexity Analysis:

• For each query, constructing an adjacency list for the subgraph might take O(V + E) space. The visited array for BFS/DFS takes O(V) space.
• Total space would be O(V + E) per query if the subgraph is rebuilt, or O(V + E) overall if the original graph is filtered on the fly.

Why it's inefficient:

Given typical constraints where N (nodes) can be up to 10^5, E (edges) up to 10^5, and Q (queries) up to 10^5, a solution of O(Q * (V + E)) would be roughly 10^5 * (10^5 + 10^5) = 2 * 10^{10} operations in the worst case. This is far too slow and would lead to a Time Limit Exceeded (TLE) error. The problem lies in repeatedly building subgraphs and running searches for each query independently. We need a way to reuse computations.

The Optimized Approach: Offline Processing with DSU

The key to solving 1697. Checking Existence of Edge Length Limited Paths efficiently lies in two powerful techniques: offline processing and the Disjoint Set Union (DSU) data structure.

Offline processing means that we don't process queries in the order they are given. Instead, we reorder them to facilitate a more efficient global strategy. For this problem, it involves sorting both the queries and the graph edges.

DSU allows us to efficiently track connected components. As we consider edges in increasing order of weight, we can use DSU to merge components, effectively building up connectivity.

How Offline Processing and DSU Work Together

The brilliant insight is this: if a path exists between p and q with a limit L, it means all edges on that path have weights < L. If we consider another query [p', q', L'] where L' > L, any path valid for L will also be valid for L'. This suggests that if we process queries in increasing order of their limit, we can progressively add edges to our DSU structure.

As we increase the limit for queries, we can gradually "activate" edges from the graph whose weights are less than the current query's limit. By adding these edges to a DSU, we merge connected components. When a query [p, q, limit] comes, all edges with weight < limit would have already been considered and potentially added to the DSU. We then simply check if p and q belong to the same connected component using DSU's find operation.

Let's break down the optimal algorithm step-by-step.

Step 1: Augment and Sort Queries

The queries are given as [p_j, q_j, limit_j]. When we sort them, we'll lose their original indices, which are necessary to store results in the correct order in the answer array. Therefore, the first step is to augment each query with its original index. So, each query [p, q, limit] becomes [limit, p, q, original_index].

After augmentation, sort the queries list in ascending order based on their limit.

Example:

Original queries: [[0, 2, 5], [0, 2, 3]] Augmented queries: [[5, 0, 2, 0], [3, 0, 2, 1]] Sorted augmented queries: [[3, 0, 2, 1], [5, 0, 2, 0]]

Step 2: Sort Graph Edges

Sort the edges list in ascending order based on their weight. So, each edge [u, v, weight] becomes [weight, u, v].

Example:

Original edges: [[0, 1, 2], [1, 2, 4]] Sorted edges: [[2, 0, 1], [4, 1, 2]]

Step 3: Initialize Disjoint Set Union (DSU)

Initialize a DSU structure for n nodes. Each node i starts as its own parent (parent[i] = i), and its component size is 1 (size[i] = 1). This setup represents n distinct components, one for each node.

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n

    def find(self, i):
        if self.parent[i] == i:
            return i
        self.parent[i] = self.find(self.parent[i])
        return self.parent[i]

    def union(self, i, j):
        root_i = self.find(i)
        root_j = self.find(j)
        if root_i != root_j:
            # Union by size/rank (here by size)
            if self.size[root_i] < self.size[root_j]:
                root_i, root_j = root_j, root_i # Swap to always attach smaller to larger
            self.parent[root_j] = root_i
            self.size[root_i] += self.size[root_j]
            return True
        return False

Step 4: Process Queries and Edges Together

This is the core of the algorithm. We will iterate through the sorted queries. For each query, we will add all eligible edges (those with weight strictly less than the query's limit) to our DSU structure.

Initialize answer = [False] * len(queries) to store the results. Also, initialize an edge_idx = 0 pointer to keep track of the current edge we are considering from the sorted edges list.

Now, iterate through the sorted_queries: For each current_query (which is [limit, p, q, original_index]):

1. Add eligible edges: While edge_idx is less than the total number of edges AND the weight of sorted_edges[edge_idx] is strictly less than current_query.limit:
   • Take the edge [weight, u, v] from sorted_edges[edge_idx].
   • Perform dsu.union(u, v). This merges the connected components of u and v if they are not already connected.
   • Increment edge_idx.
2. Check connectivity for the current query: After adding all relevant edges, check if p and q are in the same connected component using dsu.find(p) == dsu.find(q).
3. Store result: Set answer[original_index] to the result of this connectivity check.

Step 5: Return the Result Array

After processing all queries, the answer array will contain the boolean results in the original order. Return this answer array.

Detailed Example Walkthrough

Let's use a small example to trace the algorithm for 1697. Checking Existence of Edge Length Limited Paths.

Nodes: n = 4 Edges: [[0, 1, 2], [1, 2, 4], [0, 3, 3]] Queries: [[0, 2, 5], [1, 3, 2]]

Step 1: Augment and Sort Queries

Original queries: query_0 = [0, 2, 5] query_1 = [1, 3, 2]

Augmented queries: augmented_query_0 = [5, 0, 2, 0] augmented_query_1 = [2, 1, 3, 1]

Sorted augmented queries (by limit): sorted_queries = [[2, 1, 3, 1], [5, 0, 2, 0]]

Step 2: Sort Graph Edges

Original edges: edge_0 = [0, 1, 2] edge_1 = [1, 2, 4] edge_2 = [0, 3, 3]

Sorted edges (by weight): sorted_edges = [[2, 0, 1], [3, 0, 3], [4, 1, 2]]

Step 3: Initialize DSU

dsu = DSU(4) Initial state: parent = [0, 1, 2, 3] (Each node is its own parent) size = [1, 1, 1, 1] (Each component has size 1)

answer = [False, False] (for 2 queries) edge_idx = 0

Step 4: Process Queries and Edges Together

Processing sorted_queries[0] = [2, 1, 3, 1] (limit=2, p=1, q=3, original_index=1):

1. Add eligible edges:
   • Check sorted_edges[edge_idx] (which is sorted_edges[0] = [2, 0, 1]): Weight is 2. Is 2 < current_query.limit (which is 2)? No, 2 is not strictly less than 2.
   • So, no edges are added. edge_idx remains 0.
2. Check connectivity:
   • dsu.find(1) returns 1.
   • dsu.find(3) returns 3.
   • Are they equal? 1 != 3. So, False.
3. Store result: answer[1] = False. answer is now [False, False].

Processing sorted_queries[1] = [5, 0, 2, 0] (limit=5, p=0, q=2, original_index=0):

1. Add eligible edges:

   • Current edge_idx is 0.
   • Check sorted_edges[0] = [2, 0, 1]: Weight is 2. Is 2 < current_query.limit (which is 5)? Yes.
     • dsu.union(0, 1): Roots of 0 and 1 are 0 and 1. dsu.find(0) -> 0, dsu.find(1) -> 1.
     • Assuming size[0] >= size[1], parent[1] becomes 0. size[0] becomes 2.
     • Current DSU state: parent = [0, 0, 2, 3], size = [2, 1, 1, 1] (root 0 now represents {0,1}).
     • Increment edge_idx to 1.
   • Check sorted_edges[1] = [3, 0, 3]: Weight is 3. Is 3 < current_query.limit (which is 5)? Yes.
     • dsu.union(0, 3): Roots of 0 and 3 are 0 and 3. dsu.find(0) -> 0, dsu.find(3) -> 3.
     • Assuming size[0] >= size[3], parent[3] becomes 0. size[0] becomes 2+1=3.
     • Current DSU state: parent = [0, 0, 2, 0], size = [3, 1, 1, 1] (root 0 now represents {0,1,3}).
     • Increment edge_idx to 2.
   • Check sorted_edges[2] = [4, 1, 2]: Weight is 4. Is 4 < current_query.limit (which is 5)? Yes.
     • dsu.union(1, 2): Roots of 1 and 2 are 0 and 2. dsu.find(1) (which points to 0 after path compression) -> 0, dsu.find(2) -> 2.
     • Assuming size[0] >= size[2], parent[2] becomes 0. size[0] becomes 3+1=4.
     • Current DSU state: parent = [0, 0, 0, 0], size = [4, 1, 1, 1] (root 0 now represents {0,1,2,3}).
     • Increment edge_idx to 3.
   • edge_idx is now 3, which is equal to len(sorted_edges). Loop condition edge_idx < num_edges fails. Stop adding edges.

2. Check connectivity:

   • dsu.find(0) returns 0.
   • dsu.find(2) returns 0.
   • Are they equal? 0 == 0. So, True.
3. Store result: answer[0] = True. answer is now [True, False].

Step 5: Return Result

All queries processed. Return answer = [True, False].

This walkthrough demonstrates how the DSU efficiently tracks connectivity while the sorted processing ensures that only relevant edges are considered for each query's limit.

Time and Space Complexity Analysis

Understanding the performance characteristics of our optimized algorithm is crucial for competitive programming.

Time Complexity

1. Augmenting and Sorting Queries: We add an index to each query and then sort Q queries. This takes O(Q log Q) time.
2. Sorting Edges: We sort E edges. This takes O(E log E) time.
3. DSU Initialization: Initializing N parent and size arrays takes O(N) time.
4. Processing Queries and Edges:
   • We iterate through Q queries.
   • For each query, we potentially add edges. The edge_idx pointer only moves forward and makes at most E total increments across all queries. Each union operation in DSU with path compression and union by rank/size takes amortized O(alpha(N)) time.
   • For each query, we perform two find operations, taking O(alpha(N)) amortized time.
   • So, the total time for this step is O(E * alpha(N) + Q * alpha(N)), which can be simplified to O((E + Q) * alpha(N)).

Total Time Complexity: Combining these steps, the overall time complexity is O(Q log Q + E log E + (E + Q) * alpha(N)). Since alpha(N) is practically a constant (less than 5 for any realistic N, typically alpha(10^18) is 4), this is often approximated as O(Q log Q + E log E + E + Q). For typical competitive programming constraints, this is highly efficient.

Space Complexity

1. Augmented Queries: We store Q augmented queries, taking O(Q) space.
2. Sorted Edges: We store E sorted edges, taking O(E) space.
3. DSU Structure: The parent and size arrays for DSU take O(N) space.
4. Answer Array: The answer array takes O(Q) space.

Total Space Complexity: The overall space complexity is O(N + E + Q). This is generally acceptable for the given constraints.

This optimized approach effectively tackles the problem by pre-sorting data and using DSU to perform connectivity checks efficiently, avoiding redundant computations inherent in naive solutions.

Code Implementation for Checking Existence of Edge Length Limited Paths

Here's a Python implementation of the optimized solution. This includes a DSU class for clarity and the main function that orchestrates the sorting and processing steps.

class DSU:
    """
    Disjoint Set Union (DSU) data structure with path compression and union by size.
    """
    def __init__(self, n):
        # parent[i] stores the parent of node i. Initially, each node is its own parent.
        self.parent = list(range(n))
        # size[i] stores the size of the component rooted at i. Used for union by size.
        self.size = [1] * n

    def find(self, i):
        """
        Finds the representative (root) of the set that element i belongs to.
        Performs path compression to flatten the tree structure.
        """
        if self.parent[i] == i:
            return i
        # Recursively find the root and set it as the direct parent of i.
        self.parent[i] = self.find(self.parent[i])
        return self.parent[i]

    def union(self, i, j):
        """
        Merges the sets containing elements i and j.
        Uses union by size heuristic to keep the tree shallow.
        Returns True if a merge happened, False if i and j were already in the same set.
        """
        root_i = self.find(i)
        root_j = self.find(j)

        if root_i != root_j:
            # Attach the smaller tree to the root of the larger tree.
            if self.size[root_i] < self.size[root_j]:
                root_i, root_j = root_j, root_i  # Swap to ensure root_i is the larger tree's root

            self.parent[root_j] = root_i
            self.size[root_i] += self.size[root_j]
            return True
        return False

class Solution:
    def distanceLimitedPathsExist(self, n: int, edgeList: list[list[int]], queries: list[list[int]]) -> list[bool]:
        """
        Checks for existence of edge-length-limited paths in a graph.

        Args:
            n: The number of nodes in the graph.
            edgeList: A list of edges, where each edge is [u, v, weight].
            queries: A list of queries, where each query is [p, q, limit].

        Returns:
            A list of booleans, where results[i] is true if a path exists for queries[i].
        """

        # Step 1: Augment and Sort Queries
        # Each query [p, q, limit] is augmented with its original index
        # and sorted by its limit.
        # Format: [limit, p, q, original_index]
        augmented_queries = []
        for i, (p, q, limit) in enumerate(queries):
            augmented_queries.append([limit, p, q, i])
        augmented_queries.sort()

        # Step 2: Sort Graph Edges
        # Edges are sorted by their weight in ascending order.
        # Format: [weight, u, v]
        edgeList.sort(key=lambda x: x[2]) # Sort by weight (third element)

        # Step 3: Initialize Disjoint Set Union (DSU)
        dsu = DSU(n)

        # Initialize the result array with False values
        results = [False] * len(queries)

        # edge_idx will keep track of the next edge to consider from the sorted edgeList
        edge_idx = 0
        num_edges = len(edgeList)

        # Step 4: Process Queries and Edges Together
        # Iterate through the sorted queries
        for limit, p, q, original_index in augmented_queries:
            # Add all edges whose weight is strictly less than the current query's limit
            # This is where offline processing shines: we only add edges once
            # and reuse the DSU state for subsequent queries with higher limits.
            while edge_idx < num_edges and edgeList[edge_idx][2] < limit:
                u, v, _ = edgeList[edge_idx]
                dsu.union(u, v)
                edge_idx += 1

            # Step 5: Check connectivity for the current query
            # After adding all relevant edges, check if p and q are in the same component
            # by comparing their representatives (roots).
            if dsu.find(p) == dsu.find(q):
                results[original_index] = True

        # Step 6: Return the Result Array
        return results

Key Aspects of the Code

• DSU Class: A standard implementation of DSU with find (using path compression) and union (using union by size). These optimizations are critical for achieving the near-constant amortized time complexity.
• Augmenting Queries: The augmented_queries list stores the original index along with the query details. This is crucial for placing the results in the correct order after processing.
• Sorting: Both augmented_queries and edgeList are sorted according to their respective critical values (limit for queries, weight for edges).
• Two Pointers (or edge_idx): The edge_idx pointer ensures that edges are added to the DSU structure only once. For each query, it moves forward, progressively including more edges with increasing weights as the limit for queries increases. This avoids re-processing edges or rebuilding the graph for every query.
• Connectivity Check: dsu.find(p) == dsu.find(q) is the direct and efficient way to check if two nodes are connected within the current set of "active" edges (those with weights less than the query's limit).

This implementation provides a robust and efficient solution to the problem, embodying the principles of offline processing and Disjoint Set Union.

Common Mistakes and Pitfalls

Even with a clear understanding of the algorithm, it's easy to stumble upon common pitfalls when implementing problem 1697. Checking Existence of Edge Length Limited Paths. Awareness of these can save significant debugging time.

1. Incorrect Sorting Order

• Queries: Queries must be sorted by their limit in ascending order. If sorted in descending order or by any other criteria, the offline processing logic breaks down. We rely on the cumulative effect of increasing limits.
• Edges: Edges must be sorted by their weight in ascending order. If edges are not sorted, or sorted incorrectly, we won't be able to incrementally add them based on the current query's limit.

2. Forgetting Original Query Indices

When you sort the queries list, its elements change positions. If you don't store the original index of each query, you won't be able to map the computed results back to their correct positions in the final answer array. Augmenting queries with their original indices (e.g., [limit, p, q, original_index]) is a common and effective way to handle this.

3. Strict Inequality vs. Non-Strict Inequality

The problem statement explicitly says "strictly less than limit". This means an edge with weight = limit is not allowed. Pay close attention to this in your while loop condition (edgeList[edge_idx][2] < limit not edgeList[edge_idx][2] <= limit). A single character difference can lead to incorrect answers.

4. Flawed DSU Implementation

A DSU implementation without path compression and union by rank/size (or union by size) will often work for small test cases but will TLE on larger ones. The amortized O(alpha(N)) complexity relies heavily on these optimizations. Ensure your find method uses path compression and your union method uses a heuristic to keep trees shallow.

5. Off-by-One Errors with Indices

Graph problems often involve 0-indexed or 1-indexed nodes. Ensure consistency throughout your code. For n nodes, 0 to n-1 is standard in competitive programming.

6. Mismanaging the edge_idx Pointer

The edge_idx pointer in the main loop must correctly advance. It should only increment when an edge has been processed (i.e., added to the DSU). It also needs to handle the case where edge_idx reaches the end of the edgeList array. The while edge_idx < num_edges and ... condition correctly handles this.

7. Global vs. Local DSU State

Remember that the DSU state is global across queries. Edges added for a query with a smaller limit remain part of the DSU for subsequent queries with larger limits. This is the essence of offline processing combined with DSU. Do not re-initialize the DSU for each query.

By carefully reviewing these common errors, you can improve the robustness and correctness of your solution for "1697. Checking Existence of Edge Length Limited Paths".

Further Exploration and Problem Variations

While we've covered a comprehensive approach to "1697. Checking Existence of Edge Length Limited Paths", the concepts of offline processing and DSU are widely applicable. Here are some avenues for further exploration:

Other Applications of DSU

DSU is a versatile data structure used in many algorithms beyond this specific problem:

• Kruskal's Algorithm: Finding the Minimum Spanning Tree (MST) of a graph. Kruskal's algorithm fundamentally relies on sorting edges by weight and then using DSU to check if adding an edge creates a cycle, thereby efficiently constructing the MST. The core idea of incrementally adding edges based on weight and checking connectivity directly parallels our approach to "1697. Checking Existence of Edge Length Limited Paths". If you're interested in building MSTs, consider exploring Minimum Spanning Tree with Prim's Algorithm for another perspective.
• Connectivity Problems: Determining if two nodes are connected, counting connected components, or checking if adding an edge creates a cycle in various graph scenarios.
• Percolation Simulation: Modeling scenarios like water flowing through a porous medium or virus spread, where elements become "connected" over time.
• "Friend Circles" / "Number of Provinces": These are classic LeetCode problems that are directly solvable with DSU by merging sets of connected individuals.

Online vs. Offline Processing

This problem was solved using an "offline" approach. An "online" algorithm would process each query as it arrives, without knowing future queries. If edge weights could change or new edges could be added between queries, an offline approach might not be feasible, and more complex data structures like dynamic connectivity data structures (ee.g., Link-Cut Trees, dynamic graphs using segment trees or binary indexed trees over a graph) might be required. These online variations pose significantly greater challenges due to the need to efficiently update the graph structure and query connectivity in real-time. Consider what changes would make this problem "online" and how your solution would need to adapt; it's a testament to the power of offline processing when applicable.

Path Queries on Trees

For tree structures, path queries (like finding the maximum edge on a path, the k-th smallest element, or sum of weights) can often be handled efficiently using techniques like Lowest Common Ancestor (LCA) combined with binary lifting, heavy-light decomposition, or centroid decomposition. These methods preprocess the tree to answer queries quickly. While this problem involves a general graph, understanding tree-specific path queries can provide deeper insights into graph algorithms, particularly how structure can be leveraged for faster queries. For instance, finding the maximum edge on a path in a tree is directly related to the "bottleneck path" concept, which is often solved efficiently on an MST.

Edge-Weighted Graph Variations

What if, instead of strictly less than, the condition was "at most limit"? Or "at least limit"? These variations would only require minor tweaks to the sorting order or comparison operator, but they are good thought experiments to ensure full understanding. What if we wanted the path with the minimum maximum edge weight? This directly leads to concepts like bottleneck paths, which often reduce to finding paths within the Minimum Spanning Tree. Other LeetCode graph problems, such as Leetcode 1976: Number of Ways to Arrive at Destination Explained, which involves counting shortest paths, demonstrate how different constraints require distinct algorithmic approaches beyond basic connectivity.

Exploring these related problems and variations will solidify your understanding of graph algorithms, DSU, and efficient query processing, extending your skills far beyond 1697. Checking Existence of Edge Length Limited Paths.

Frequently Asked Questions

Q: What is the main challenge of LeetCode problem 1697, Checking Existence of Edge Length Limited Paths?

A: The core challenge is efficiently checking connectivity between nodes while adhering to a strict edge length limit for each path. Naive BFS/DFS for every query is too slow for large graphs and numerous queries, necessitating a more optimized approach.

Q: Why is the Disjoint Set Union (DSU) data structure crucial for solving this problem?

A: DSU efficiently manages connected components. By integrating edges into the DSU structure as their weights become eligible (i.e., less than a query's limit), we can quickly determine if two nodes are in the same component, indicating a valid path.

Q: How does offline processing improve the efficiency of solving this problem?

A: Offline processing involves sorting both graph edges by weight and queries by their limit. This allows us to process queries in a way that incrementally builds connectivity with DSU, reusing computations and avoiding repeated subgraph constructions or traversals for each query.

Conclusion

Tackling 1697. Checking Existence of Edge Length Limited Paths serves as an excellent benchmark for proficiency in intermediate to advanced graph algorithms. We've explored how a naive, brute-force approach quickly succumbs to performance limitations due to redundant computations. The elegant and efficient solution hinges on two core principles: offline processing and the Disjoint Set Union (DSU) data structure.

By augmenting queries with their original indices and sorting both queries (by limit) and edges (by weight), we enable a single, progressive pass over the edges. This allows us to incrementally build up connected components using DSU. For each query, we incorporate all edges whose weights fall below the current limit into our DSU structure and then perform a quick find operation to check connectivity. This strategy avoids repeatedly scanning the graph for each query, leading to an optimal time complexity of O(Q log Q + E log E + (E + Q) * alpha(N)) and space complexity of O(N + E + Q).

Mastering this problem reinforces crucial algorithmic thinking: identifying opportunities for pre-processing, understanding how to leverage data structures like DSU for efficient connectivity checks, and recognizing when offline processing can transform an intractable problem into an efficient one. The techniques learned here are broadly applicable across competitive programming and real-world software development scenarios involving graph analysis. Keep practicing similar problems to solidify these powerful concepts!

Further Reading & Resources

• Disjoint-set data structure - Wikipedia
• LeetCode 1697: Checking Existence of Edge Length Limited Paths
• Disjoint Set Union (Union by Rank and Path Compression) - GeeksforGeeks