Shortest Distance After Road Addition Queries I
Time O(n^2) · Space O(n^2) · Official statement on LeetCode
Solutions
// Time: O(n^2)
// Space: O(n^2)
// bfs
class Solution {
public:
vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
vector<vector<int>> adj(n);
for (int u = 0; u + 1 < n; ++u) {
adj[u].emplace_back(u + 1);
}
vector<int> dist(n);
iota(begin(dist), end(dist), 0);
const auto& bfs = [&](int u, int v) {
adj[u].emplace_back(v);
vector<int> q = {u};
while (!empty(q)) {
vector<int> new_q;
for (const auto& u : q) {
for (const auto& v : adj[u]) {
if (dist[u] + 1 >= dist[v]) {
continue;
}
dist[v] = dist[u] + 1;
new_q.emplace_back(v);
}
}
q = move(new_q);
}
return dist[n - 1];
};
vector<int> result;
result.reserve(n);
for (const auto& q : queries) {
result.emplace_back(bfs(q[0], q[1]));
}
return result;
}
};
// Time: O(n^2 * logn)
// Space: O(n^2)
// dijkstra's algorithm
class Solution2 {
public:
vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
vector<vector<pair<int, int>>> adj(n);
for (int u = 0; u + 1 < n; ++u) {
adj[u].emplace_back(u + 1, 1);
}
vector<int> dist(n);
iota(begin(dist), end(dist), 0);
const auto& dijkstra = [&](int u, int v) {
adj[u].emplace_back(v, 1);
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> min_heap;
min_heap.emplace(dist[u], u);
while (!empty(min_heap)) {
const auto [curr, u] = min_heap.top(); min_heap.pop();
if (curr > dist[u]) {
continue;
}
for (const auto& [v, w] : adj[u]) {
if (curr + w >= dist[v]) {
continue;
}
dist[v] = curr + w;
min_heap.emplace(dist[v], v);
}
}
return dist[n - 1];
};
vector<int> result;
result.reserve(n);
for (const auto& q : queries) {
result.emplace_back(dijkstra(q[0], q[1]));
}
return result;
}
};
Beginner Explanation
What is Shortest Distance After Road Addition Queries I?
Shortest Distance After Road Addition Queries I (LeetCode #3243) is a Medium problem that primarily trains graph.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with graph, dijkstras algorithm, and queue bfs.
- Only then translate the idea into code.
Why this problem matters
It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: Graph, Dijkstra's Algorithm, BFS.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Shortest Distance After Road Addition Queries I
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to graph, dijkstras algorithm, and queue bfs.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(n^2)) and space (O(n^2)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(n^2) time and O(n^2) space.
Pattern focus: graph, dijkstras algorithm, and queue bfs
Use the pattern as a checklist:
- graph — confirm the invariant holds after each step
- dijkstras algorithm — confirm the invariant holds after each step
- queue bfs — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(n^2) |
| Space | O(n^2) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Shortest Distance After Road Addition Queries I
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for graph, dijkstras algorithm, and queue bfs — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to graph, dijkstras algorithm, and queue bfs:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: graph.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Shortest Distance After Road Addition Queries I in a second language (cpp, python).
- Drill 3–5 more problems tagged graph.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the graph, dijkstras algorithm, and queue bfs approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Shortest Distance After Road Addition Queries I (#3243) — Medium. Pattern: graph, dijkstras algorithm, and queue bfs. Complexity: O(n^2) time / O(n^2) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Shortest Distance After Road Addition Queries I?+
The reference solutions aim for O(n^2) time and O(n^2) space. Always re-derive complexity from the code you write in the interview.
What pattern does Shortest Distance After Road Addition Queries I use?+
It primarily maps to graph, dijkstras algorithm, and queue bfs, within the broader topic of graph.
Is Shortest Distance After Road Addition Queries I good for interviews?+
Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/shortest-distance-after-road-addition-queries-i/