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#3526HardPremium on LC~50 min

Range XOR Queries with Subarray Reversals

Time O(n + qlogn) · Space O(n) · Official statement on LeetCode

binary search treetree traversal
cpppython

Solutions

// Time:  O(n + qlogn)
// Space: O(n)

// template from: https://cp-algorithms.com/data_structures/treap.html
using pTreapNode = struct TreapNode *;

struct TreapNode {
    int prior, value, cnt;
    int xor_sum;  // added
    bool rev;
    pTreapNode l, r;

    TreapNode (int v) : value(v), prior(rand()), cnt(1), xor_sum(v), rev(false), l(nullptr), r(nullptr) {}  // added
};

int cnt(pTreapNode t) {
    return t ? t->cnt : 0;
}

int xor_sum(pTreapNode t) {  // added
    return t ? t->xor_sum : 0;
}

void upd_cnt(pTreapNode t) {
    if (t) {
        t->cnt = cnt(t->l) + cnt(t->r) + 1;
        t->xor_sum = t->value ^ xor_sum(t->l) ^ xor_sum(t->r);  // added
    }
}

void push(pTreapNode t) {
    if (t && t->rev) {
        t->rev = false;
        swap(t->l, t->r);
        if (t->l) t->l->rev ^= true;
        if (t->r) t->r->rev ^= true;
    }
}

void merge(pTreapNode& t, pTreapNode l, pTreapNode r) {
    push(l);
    push(r);
    if (!l || !r)
        t = l ? l : r;
    else if (l->prior > r->prior)
        merge(l->r, l->r, r), t = l;
    else
        merge(r->l, l, r->l), t = r;
    upd_cnt(t);
}

void split(pTreapNode t, pTreapNode& l, pTreapNode& r, int key, int add = 0) {
    if (!t)
        return void(l = r = 0);
    push(t);
    int cur_key = add + cnt(t->l);
    if (key <= cur_key)
        split(t->l, l, t->l, key, add), r = t;
    else
        split(t->r, t->r, r, key, add + 1 + cnt(t->l)), l = t;
    upd_cnt(t);
}

void reverse(pTreapNode t, int l, int r) {
    pTreapNode t1, t2, t3;
    split(t, t1, t2, l);
    split(t2, t2, t3, r - l + 1);
    t2->rev ^= true;
    merge(t, t1, t2);
    merge(t, t, t3);
}

void heapify(pTreapNode t) {
    if (!t) return;
    pTreapNode max = t;
    if (t->l && t->l->prior > max->prior)
        max = t->l;
    if (t->r && t->r->prior > max->prior)
        max = t->r;
    if (max != t) {
        swap(t->prior, max->prior);
        heapify(max);
    }
}

pTreapNode build(const vector<int>& a, int i, int n) {
    if (n == 0) return nullptr;
    int mid = n / 2;
    auto t = new TreapNode(a[i + mid]);
    t->l = build(a, i, mid);
    t->r = build(a, i + mid + 1, n - mid - 1);
    heapify(t);
    upd_cnt(t);
    return t;
}

// treap
class Solution {
public:
    vector<int> getResults(vector<int>& nums, vector<vector<int>>& queries) {
        pTreapNode root = build(nums, 0, size(nums));
        const auto& update = [&](int index, int value) {
            pTreapNode left, mid, right;
            split(root, left, mid, index);
            split(mid, mid, right, 1);
            mid->value = value;
            upd_cnt(mid);
            merge(root, left, mid);
            merge(root, root, right);
        };

        const auto& query = [&](int left, int right) {
            pTreapNode t1, t2, t3;
            split(root, t1, t2, left);
            split(t2, t2, t3, right - left + 1);
            int result = xor_sum(t2);
            merge(root, t1, t2);
            merge(root, root, t3);
            return result;
        };

        vector<int> result;
        for (const auto& q : queries) {
            if (q[0] == 1) {
                update(q[1], q[2]);
            } else if (q[0] == 2) {
                result.emplace_back(query(q[1], q[2]));
            } else if (q[0] == 3) {
                reverse(root, q[1], q[2]);
            }
        }
        return result;
    }
};

Beginner Explanation

What is Range XOR Queries with Subarray Reversals?

Range XOR Queries with Subarray Reversals (LeetCode #3526) is a Hard problem that primarily trains binary search tree.

How to think about it

  1. Restate the goal in your own words before coding.
  2. Work a tiny example by hand so the invariant becomes obvious.
  3. Identify the pattern — this problem aligns with tree traversal.
  4. 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: Treap.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Range XOR Queries with Subarray Reversals

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 tree traversal.

Core solution narrative

  1. Define the state you track (pointers, DP cell, set membership, stack top, etc.).
  2. Explain the transition when you process the next element.
  3. Call out time (O(n + qlogn)) and space (O(n)) before coding.
  4. 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 + qlogn) time and O(n) space.

Pattern focus: tree traversal

Use the pattern as a checklist:

  • tree traversal — 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 + qlogn)
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 Range XOR Queries with Subarray Reversals

  1. Skipping edge cases — empty collections, single-element inputs, max constraints.
  2. Wrong invariant for tree traversal — updating state too early or too late.
  3. Mutating input unexpectedly when the problem forbids it.
  4. Off-by-one in windows, ranges, or binary search bounds.
  5. Ignoring overflow / precision for integer arithmetic problems.
  6. Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

  1. Primary optimized path — best complexity for typical interviews.
  2. 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 tree traversal:

  • 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: binary search tree.

Follow-up Interview Questions

Follow-ups

  1. How does the solution change if the input is a stream?
  2. Can you solve it in-place?
  3. What if duplicates must be handled differently?
  4. How would you parallelize the approach?
  5. Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

  1. Re-solve Range XOR Queries with Subarray Reversals in a second language (cpp, python).
  2. Drill 3–5 more problems tagged binary search tree.
  3. Teach the solution out loud in under 5 minutes.
  4. Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Range XOR Queries with Subarray Reversals: show input structure (binary search tree), highlight the moving parts of the tree traversal 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 tree traversal 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

Range XOR Queries with Subarray Reversals (#3526) — Hard. Pattern: tree traversal. Complexity: O(n + qlogn) time / O(n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Range XOR Queries with Subarray Reversals?+

The reference solutions aim for O(n + qlogn) time and O(n) space. Always re-derive complexity from the code you write in the interview.

What pattern does Range XOR Queries with Subarray Reversals use?+

It primarily maps to tree traversal, within the broader topic of binary search tree.

Is Range XOR Queries with Subarray Reversals 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/range-xor-queries-with-subarray-reversals/