Problem Description

You are given a connected, undirected, weighted graph with n nodes (labeled from 0 to n-1) and a list of edges. Each edge is represented as [u, v, w], meaning there's an edge between nodes u and v with weight w.

You need to find the shortest path from a starting node s to a destination node d, but with a special power: you can "hop over" (make the weight 0) for at most k edges along your path.

Key Points:

• The graph has n nodes (0-indexed)
• Each edge in edges is bidirectional with format [u, v, w] where w is the weight
• You want to go from node s to node d
• You can choose up to k edges and treat their weights as 0 (effectively "hopping" over them for free)
• Return the minimum total path weight from s to d

Example Understanding: If you have a path from s to d that uses edges with weights [5, 3, 7, 2] and k=2, you could choose to hop over the two heaviest edges (7 and 5), making your total cost only 3 + 2 = 5 instead of 5 + 3 + 7 + 2 = 17.

The challenge is to strategically choose which edges to hop over to minimize the total path weight from source to destination.

Intuition

The problem asks us to find the shortest path with the ability to skip up to k edges. This is a variation of the classic shortest path problem with an additional dimension - the number of "hops" we've used.

Think of it this way: at each node, we don't just care about "what's the shortest distance to reach here?" but rather "what's the shortest distance to reach here having used exactly 0 hops, 1 hop, 2 hops, ..., up to k hops?"

This leads us to track state in two dimensions:

• Which node we're at
• How many hops we've used so far

For each node, we maintain k+1 different distances: dist[node][hops_used] represents the minimum distance to reach that node using exactly hops_used free edges.

When we're at a node with some number of hops already used, we have two choices for each neighboring edge:

1. Use a hop: Skip this edge (treat its weight as 0) if we haven't used all k hops yet. This moves us to the neighbor with one more hop used but no additional distance.
2. Pay the cost: Travel the edge normally, adding its weight to our distance but keeping the same hop count.

This naturally fits into Dijkstra's algorithm framework, but instead of relaxing just distances, we're relaxing distance-hop pairs. We use a priority queue with entries (distance, node, hops_used) and always process the smallest distance first.

The key insight is that we're essentially running Dijkstra on an expanded graph where each original node is split into k+1 copies (one for each possible hop count), and we're finding the shortest path in this expanded state space.

Finally, the answer is the minimum among all dist[d][0], dist[d][1], ..., dist[d][k] since we can reach the destination using any number of hops from 0 to k.

Learn more about Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The solution implements a modified Dijkstra's algorithm to handle the additional dimension of hop counts.

Step 1: Build the Graph

g = [[] for _ in range(n)]
for u, v, w in edges:
    g[u].append((v, w))
    g[v].append((u, w))

We create an adjacency list representation where g[u] contains all neighbors of node u along with their edge weights. Since the graph is undirected, we add edges in both directions.

Step 2: Initialize Distance Matrix

dist = [[inf] * (k + 1) for _ in range(n)]
dist[s][0] = 0

We create a 2D distance array where dist[u][t] represents the minimum distance to reach node u using exactly t hops. Initially, all distances are set to infinity except dist[s][0] = 0 (we start at source s with 0 hops used and 0 distance).

Step 3: Priority Queue Setup

pq = [(0, s, 0)]  # (distance, node, hops_used)

We use a min-heap priority queue that stores tuples of (current_distance, current_node, hops_used). This ensures we always process the state with minimum distance first.

Step 4: Modified Dijkstra's Algorithm

while pq:
    dis, u, t = heappop(pq)
    for v, w in g[u]:
        # Option 1: Use a hop (make edge weight 0)
        if t + 1 <= k and dist[v][t + 1] > dis:
            dist[v][t + 1] = dis
            heappush(pq, (dis, v, t + 1))
      
        # Option 2: Pay the edge cost
        if dist[v][t] > dis + w:
            dist[v][t] = dis + w
            heappush(pq, (dis + w, v, t))

For each state (dis, u, t) popped from the queue, we explore all neighbors v of node u:

• Using a hop: If we haven't used all k hops (t + 1 <= k), we can reach neighbor v with the same distance dis but using one more hop. We update dist[v][t + 1] if this gives a better path.

• Paying the cost: We can reach neighbor v by paying the edge weight w, keeping the same hop count t. We update dist[v][t] if dis + w is better than the current known distance.

Step 5: Find the Answer

return int(min(dist[d]))

The shortest path to destination d is the minimum among all possible hop counts: min(dist[d][0], dist[d][1], ..., dist[d][k]).

Time Complexity: O((n * k) * log(n * k) * m) where m is the number of edges, as we potentially process each state (node, hops) and each edge from that state.

Space Complexity: O(n * k) for the distance matrix and priority queue.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given:

• Graph with 4 nodes (0, 1, 2, 3)
• Edges: [[0,1,5], [1,2,3], [0,2,9], [2,3,1]]
• Start: s = 0, Destination: d = 3
• Maximum hops: k = 1

Graph visualization:

    0 ---5--- 1
    |         |
    9         3
    |         |
    2 ---1--- 3

Step 1: Initialize

• Build adjacency list:

  • g[0] = [(1,5), (2,9)]
  • g[1] = [(0,5), (2,3)]
  • g[2] = [(0,9), (1,3), (3,1)]
  • g[3] = [(2,1)]

• Initialize distance matrix dist[node][hops_used]:

  dist[0][0] = 0, dist[0][1] = inf
  dist[1][0] = inf, dist[1][1] = inf
  dist[2][0] = inf, dist[2][1] = inf
  dist[3][0] = inf, dist[3][1] = inf

• Priority queue: [(0, 0, 0)] (distance=0, node=0, hops=0)

Step 2: Process from node 0 with 0 hops used

• Pop (0, 0, 0) from queue
• Explore neighbors:
  • To node 1 (edge weight 5):
    • Option 1 (use hop): dist[1][1] = 0 (better than inf)
    • Option 2 (pay cost): dist[1][0] = 5 (better than inf)
    • Add to queue: (0, 1, 1) and (5, 1, 0)
  • To node 2 (edge weight 9):
    • Option 1 (use hop): dist[2][1] = 0 (better than inf)
    • Option 2 (pay cost): dist[2][0] = 9 (better than inf)
    • Add to queue: (0, 2, 1) and (9, 2, 0)

Step 3: Process minimum distance state (0, 1, 1)

• Pop (0, 1, 1) (reached node 1 with distance 0 using 1 hop)
• Explore neighbors:
  • To node 0: Skip (no improvement)
  • To node 2 (edge weight 3):
    • Cannot use another hop (already used k=1)
    • Option 2 (pay cost): dist[2][1] = 3 (0+3=3, worse than current 0)

Step 4: Process (0, 2, 1)

• Pop (0, 2, 1) (reached node 2 with distance 0 using 1 hop)
• Explore neighbors:
  • To node 3 (edge weight 1):
    • Cannot use another hop (already used k=1)
    • Option 2 (pay cost): dist[3][1] = 1 (0+1=1, better than inf)
    • Add to queue: (1, 3, 1)

Step 5: Process (1, 3, 1)

• Pop (1, 3, 1) - we've reached destination!
• Continue processing remaining states...

Step 6: Process (5, 1, 0)

• Pop (5, 1, 0) (reached node 1 with distance 5 using 0 hops)
• To node 2 (edge weight 3):
  • Option 1 (use hop): dist[2][1] = 5 (worse than current 0)
  • Option 2 (pay cost): dist[2][0] = 8 (5+3=8, better than 9)
  • Add to queue: (8, 2, 0)

Step 7: Process (8, 2, 0)

• Pop (8, 2, 0)
• To node 3 (edge weight 1):
  • Option 1 (use hop): dist[3][1] = 8 (worse than current 1)
  • Option 2 (pay cost): dist[3][0] = 9 (8+1=9, better than inf)
  • Add to queue: (9, 3, 0)

Final Result:

• dist[3][0] = 9 (path: 0→1→2→3 with weights 5+3+1)
• dist[3][1] = 1 (path: 0→2→3 with hop on 0→2, paying only for 2→3)
• Answer: min(9, 1) = 1

The optimal path uses the direct expensive edge 0→2 as a free hop, then pays only 1 for edge 2→3.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * k * (n + m) * log(n * k)) which simplifies to O(n^2 * k * log(n * k)) in the worst case when m = O(n^2).

The algorithm uses a modified Dijkstra's approach with an additional dimension for the number of hops. Each node can be visited up to k + 1 times (for different hop counts from 0 to k). The priority queue can contain at most n * (k + 1) states. Each state extraction takes O(log(n * k)) time. For each extracted state, we explore all adjacent edges. In the worst case with a dense graph where m = O(n^2), we might push O(n * k) states to the queue, each taking O(log(n * k)) time. However, the reference answer suggests O(n^2 * log n), which appears to be considering the case where k is treated as a constant or when the graph is sparse and the dominant factor is n^2 comparisons across all possible paths.

Space Complexity: O(n * k)

The space is dominated by:

• The dist array: O(n * (k + 1)) = O(n * k)
• The adjacency list g: O(n + m) where m is the number of edges
• The priority queue can hold at most O(n * k) elements

Since the dist array is the dominant factor, the overall space complexity is O(n * k), where n represents the number of nodes and k represents the maximum number of free hops allowed.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Checking if Current State is Already Processed

The current implementation may process the same state (node, hops_used) multiple times if a better path to that state is found later. This happens because we don't check if the current distance from the priority queue is outdated.

Problem: When we pop (dis, u, t) from the priority queue, dis might be larger than dist[u][t] if we've already found a better path to state (u, t) after this entry was added to the queue.

Solution: Add a check to skip processing if we've already found a better path:

while priority_queue:
    current_dist, current_node, hops_used = heappop(priority_queue)
  
    # Skip if we've already found a better path to this state
    if current_dist > dist[current_node][hops_used]:
        continue
  
    # Rest of the code remains the same...

2. Inefficient Hop Usage Strategy

A subtle pitfall is that the algorithm explores all possibilities but doesn't guarantee optimal hop usage early on. We might use hops on small-weight edges early in the path when saving them for heavier edges later would be better.

Why this isn't actually a problem: The algorithm explores ALL possible combinations of hop usage by maintaining separate distances for each hop count. The state space (node, hops_used) ensures we consider all strategies.

3. Memory Limit Issues with Large Graphs

For very large values of n and k, the 2D distance array dist[n][k+1] can consume significant memory.

Solution: Use a dictionary to store only visited states:

from collections import defaultdict

dist = defaultdict(lambda: inf)
dist[(s, 0)] = 0
priority_queue = [(0, s, 0)]

while priority_queue:
    current_dist, current_node, hops_used = heappop(priority_queue)
  
    if current_dist > dist[(current_node, hops_used)]:
        continue
  
    for neighbor, edge_weight in graph[current_node]:
        # Option 1: Use a hop
        if hops_used + 1 <= k and dist[(neighbor, hops_used + 1)] > current_dist:
            dist[(neighbor, hops_used + 1)] = current_dist
            heappush(priority_queue, (current_dist, neighbor, hops_used + 1))
      
        # Option 2: Pay the cost
        if dist[(neighbor, hops_used)] > current_dist + edge_weight:
            dist[(neighbor, hops_used)] = current_dist + edge_weight
            heappush(priority_queue, (current_dist + edge_weight, neighbor, hops_used))

# Find answer
return min(dist[(d, i)] for i in range(k + 1))

4. Edge Case: Unreachable Destination

If the destination d is not reachable from source s, all values in dist[d] will remain inf.

Solution: Add a check before returning:

result = min(dist[d])
return -1 if result == inf else int(result)

5. Integer Overflow with Float Infinity

Using float('inf') and then converting to int can cause issues in some systems.

Solution: Use a large integer constant instead:

INF = 10**9  # or sys.maxsize
dist = [[INF] * (k + 1) for _ in range(n)]

The most critical pitfall is #1 (not checking for already processed states), which can significantly impact performance on dense graphs with many paths between nodes.