#2322Hard~50 min
Minimum Score After Removals on a Tree
Time O(n^2) · Space O(n) · Official statement on LeetCode
Solutions
// Time: O(n^2)
// Space: O(n)
// dfs with stack
class Solution {
public:
int minimumScore(vector<int>& nums, vector<vector<int>>& edges) {
vector<vector<int>> adj(size(nums));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
vector<int> left(size(nums)), right(size(nums));
const auto& is_ancestor = [&](int a, int b) {
return left[a] <= left[b] && right[b] <= right[a];
};
const auto& iter_dfs = [&]() {
int cnt = 0;
vector<tuple<int, int, int>> stk;
stk.emplace_back(1, 0, -1);
while (!empty(stk)) {
const auto [step, u, p] = stk.back(); stk.pop_back();
if (step == 1) {
left[u] = cnt++;
stk.emplace_back(2, u, p);
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
stk.emplace_back(1, v, u);
}
} else if (step == 2) {
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
nums[u] ^= nums[v];
}
right[u] = cnt;
}
}
};
iter_dfs();
int result = numeric_limits<int>::max();
for (int i = 1; i < size(nums); ++i) {
for (int j = i + 1; j < size(nums); ++j) {
int a, b, c;
if (is_ancestor(i, j)) {
tie(a, b, c) = tuple(nums[0] ^ nums[i], nums[i] ^ nums[j], nums[j]);
} else if (is_ancestor(j, i)) {
tie(a, b, c) = tuple(nums[0] ^ nums[j], nums[j] ^ nums[i], nums[i]);
} else {
tie(a, b, c) = tuple(nums[0] ^ nums[i] ^ nums[j], nums[i], nums[j]);
}
result = min(result, max({a, b, c}) - min({a, b, c}));
}
}
return result;
}
};
// Time: O(n^2)
// Space: O(n)
// dfs with recursion
class Solution2 {
public:
int minimumScore(vector<int>& nums, vector<vector<int>>& edges) {
vector<vector<int>> adj(size(nums));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
int cnt = 0;
vector<int> left(size(nums)), right(size(nums));
const auto& is_ancestor = [&](int a, int b) {
return left[a] <= left[b] && right[b] <= right[a];
};
function<void(int, int)> dfs = [&](int u, int p) {
left[u] = cnt++;
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
dfs(v, u);
nums[u] ^= nums[v];
}
right[u] = cnt;
};
dfs(0, -1);
int result = numeric_limits<int>::max();
for (int i = 1; i < size(nums); ++i) {
for (int j = i + 1; j < size(nums); ++j) {
int a, b, c;
if (is_ancestor(i, j)) {
tie(a, b, c) = tuple(nums[0] ^ nums[i], nums[i] ^ nums[j], nums[j]);
} else if (is_ancestor(j, i)) {
tie(a, b, c) = tuple(nums[0] ^ nums[j], nums[j] ^ nums[i], nums[i]);
} else {
tie(a, b, c) = tuple(nums[0] ^ nums[i] ^ nums[j], nums[i], nums[j]);
}
result = min(result, max({a, b, c}) - min({a, b, c}));
}
}
return result;
}
};
// Time: O(n^2)
// Space: O(n)
// dfs with recursion
class Solution3 {
public:
int minimumScore(vector<int>& nums, vector<vector<int>>& edges) {
vector<vector<int>> adj(size(nums));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
function<int(int, int, vector<int> *)> dfs = [&](int u, int p, vector<int> *result) {
int total = nums[u];
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
total ^= dfs(v, u, result);
}
result->emplace_back(total);
return total;
};
const int total = accumulate(cbegin(nums), cend(nums), 0,
[](int x, int y) {
return x ^ y;
});
int result = numeric_limits<int>::max();
for (const auto& e : edges) {
vector<vector<int>> xors(2);
dfs(e[0], e[1], &xors[0]);
dfs(e[1], e[0], &xors[1]);
for (auto& candidates : xors) {
const int total2 = candidates.back(); candidates.pop_back();
for (const auto& x : candidates) {
const auto& [a, b, c] = tuple(total ^ total2, x, total2 ^ x);
result = min(result, max({a, b, c}) - min({a, b, c}));
}
}
}
return result;
}
};
// Time: O(n^2)
// Space: O(n)
// dfs with stack
class Solution4_TLE {
public:
int minimumScore(vector<int>& nums, vector<vector<int>>& edges) {
vector<vector<int>> adj(size(nums));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
const auto& iter_dfs = [&](int u, int p, vector<int> *result) {
using RET = int;
RET ret = 0;
vector<tuple<int, int, int, shared_ptr<vector<shared_ptr<RET>>>, RET *>> stk;
stk.emplace_back(1, u, p, nullptr, &ret);
while (!empty(stk)) {
auto [step, u, p, new_rets, ret] = stk.back(); stk.pop_back();
if (step == 1) {
auto new_rets = make_shared<vector<shared_ptr<RET>>>();
stk.emplace_back(2, u, p, new_rets, ret);
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
new_rets->emplace_back(make_shared<RET>());
stk.emplace_back(1, v, u, nullptr, new_rets->back().get());
}
} else if (step == 2) {
*ret = nums[u];
for (const auto& x : *new_rets) {
*ret ^= *x;
}
result->emplace_back(*ret);
}
}
};
const int total = accumulate(cbegin(nums), cend(nums), 0,
[](int x, int y) {
return x ^ y;
});
int result = numeric_limits<int>::max();
for (const auto& e : edges) {
vector<vector<int>> xors(2);
iter_dfs(e[0], e[1], &xors[0]);
iter_dfs(e[1], e[0], &xors[1]);
for (auto& candidates : xors) {
const int total2 = candidates.back(); candidates.pop_back();
for (const auto& x : candidates) {
const auto& [a, b, c] = tuple(total ^ total2, x, total2 ^ x);
result = min(result, max({a, b, c}) - min({a, b, c}));
}
}
}
return result;
}
};
Beginner Explanation
What is Minimum Score After Removals on a Tree?
Minimum Score After Removals on a Tree (LeetCode #2322) is a Hard problem that primarily trains depth first search.
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 dfs backtracking and tree.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DFS, Tree.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Minimum Score After Removals on a Tree
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 dfs backtracking and tree.
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)) 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) space.
Pattern focus: dfs backtracking and tree
Use the pattern as a checklist:
- dfs backtracking — confirm the invariant holds after each step
- tree — 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) |
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 Minimum Score After Removals on a Tree
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dfs backtracking and tree — 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 dfs backtracking and tree:
- 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: depth first search.
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 Minimum Score After Removals on a Tree in a second language (cpp, python).
- Drill 3–5 more problems tagged depth first search.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Conceptual diagram for Minimum Score After Removals on a Tree: show input structure (depth first search), highlight the moving parts of the dfs backtracking and tree approach, and annotate each step with the maintained invariant and complexity.
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dfs backtracking and tree 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
Minimum Score After Removals on a Tree (#2322) — Hard. Pattern: dfs backtracking and tree. Complexity: O(n^2) time / O(n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Minimum Score After Removals on a Tree?+
The reference solutions aim for O(n^2) time and O(n) space. Always re-derive complexity from the code you write in the interview.
What pattern does Minimum Score After Removals on a Tree use?+
It primarily maps to dfs backtracking and tree, within the broader topic of depth first search.
Is Minimum Score After Removals on a Tree good for interviews?+
Yes — as a Hard 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/minimum-score-after-removals-on-a-tree/