#2613HardPremium on LC~50 min

Beautiful Pairs

Time O(n) on average · Space O(n) · Official statement on LeetCode

recursion random algorithms merge sort segment tree

Solutions

// Time:  O(n) on average
// Space: O(n)

// random algorithms, variant of closest pair
// reference: https://github.com/jilljenn/tryalgo/blob/master/tryalgo/closest_points.py
class Solution {
private:
    template<typename T>
    struct VectorHash {
        size_t operator()(const std::vector<T>& v) const {
            size_t seed = 0;
            for (const auto& i : v) {
                seed ^= std::hash<T>{}(i)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            }
            return seed;
        }
    };

public:
    vector<int> beautifulPair(vector<int>& nums1, vector<int>& nums2) {
        static const int INF = numeric_limits<int>::max();
        srand(time(0));

        vector<vector<int>> points;
        for (int i = 0; i < size(nums1); ++i) {
            points.push_back({nums1[i], nums2[i], i});
        }
        vector<int> result(3, INF);
        unordered_map<vector<int>, int, VectorHash<int>> lookup;
        for (int i = size(points) - 1; i >= 0; --i) {
            const auto& key = vector<int>{points[i][0], points[i][1]};
            if (lookup.count(key)) {
                result = min(result, vector<int>{0, i, lookup[key]});
            }
            lookup[key] = i;
        }
        if (result[0] == 0) {
            return {result[1], result[2]};
        }
        vector<int> order(size(points));
        iota(begin(order), end(order), 0);
        sort(begin(order), end(order), [&](const auto& a, const auto& b) {
            return points[a] < points[b];
        });
        const auto& dist = [&](auto a, auto b) {
            if (a[2] > b[2]) {
                swap(a, b);
            }
            return vector<int>{abs(a[0] - b[0]) + abs(a[1] - b[1]), a[2], b[2]};
        };

        const auto& improve = [&]() {
            unordered_map<vector<int>, vector<int>, VectorHash<int>> lookup;
            for (const auto& p : points) {
                const double size = result[0] / 2.0;
                const int i = floor(p[0] / size), j = floor(p[1] / size);
                for (int ni = i - 2; ni <= i + 2; ++ni) {
                    for (int nj = j - 2; nj <= j + 2; ++nj) {
                        const auto& key = vector<int>{ni, nj};
                        if (!lookup.count(key)) {
                            continue;
                        }
                        const auto& d = dist(p, lookup[key]);
                        if (d < result) {
                            result = d;
                            return true;
                        }
                    }
                }
                lookup[vector<int>{i, j}] = p;
            }
            return false;
        };

        random_shuffle(begin(points), end(points));
        result = dist(points[0], points[1]);
        while (improve());
        return {result[1], result[2]};
    }
};

// Time:  O(nlogn)
// Space: O(n)
// divide and conquer, merge sort, variant of closest pair
// reference: https://www.baeldung.com/cs/minimal-manhattan-distance
class Solution2 {
private:
    template<typename T>
    struct VectorHash {
        size_t operator()(const std::vector<T>& v) const {
            size_t seed = 0;
            for (const auto& i : v) {
                seed ^= std::hash<T>{}(i)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            }
            return seed;
        }
    };

public:
    vector<int> beautifulPair(vector<int>& nums1, vector<int>& nums2) {
        static const int INF = numeric_limits<int>::max();
        static const int MAX_NEIGHBOR_COUNT = (8 + 2) / 2;

        vector<vector<int>> points;
        for (int i = 0; i < size(nums1); ++i) {
            points.push_back({nums1[i], nums2[i]});
        }
        vector<int> result(3, INF);
        unordered_map<vector<int>, int, VectorHash<int>> lookup;
        for (int i = size(points) - 1; i >= 0; --i) {
            if (lookup.count(points[i])) {
                result = min(result, vector<int>{0, i, lookup[points[i]]});
            }
            lookup[points[i]] = i;
        }
        if (result[0] == 0) {
            return {result[1], result[2]};
        }
        vector<int> order(size(points));
        iota(begin(order), end(order), 0);
        sort(begin(order), end(order), [&](const auto& a, const auto& b) {
            return points[a][0] < points[b][0];
        });
        const auto& dist = [&](auto a, auto b) {
            if (a > b) {
                swap(a, b);
            }
            return vector<int>{abs(points[a][0] - points[b][0]) + abs(points[a][1] - points[b][1]), a, b};
        };

        const function<void (int, int)> merge_sort = [&](int left, int right) {
            const auto& update = [&](const auto& arr, int i) {  // added
                for (int j = size(arr) - 1; j >= 0; --j) {
                    if (points[i][1] - points[arr[j]][1] > result[0]) {
                        break;
                    }
                    assert((size(arr) - 1) - j + 1 <= MAX_NEIGHBOR_COUNT);
                    result = min(result, dist(i, arr[j]));
                }
            };
            
            if (left == right) {
                return;
            }
            const int mid = left + (right - left) / 2;
            const int x = points[order[mid]][0];  // added
            merge_sort(left, mid);
            merge_sort(mid + 1, right);
            vector<int> tmp, tmp_l, tmp_r;
            for (int l = left, r = mid + 1; l < mid + 1 || r < right + 1;) {
                if (r >= right + 1 || (l < mid + 1 && points[order[l]][1] <= points[order[r]][1])) {  // modified
                    update(tmp_r, order[l]);
                    if (x - points[order[l]][0] <= result[0]) {  // added
                        tmp_l.emplace_back(order[l]);
                    }
                    tmp.emplace_back(order[l++]);
                } else {
                    update(tmp_l, order[r]);
                    if (points[order[r]][0] - x <= result[0]) {  // added
                        tmp_r.emplace_back(order[r]);
                    }
                    tmp.emplace_back(order[r++]);
                }
            }
            copy(cbegin(tmp), cend(tmp), begin(order) + left);
        };
        
        merge_sort(0, size(points) - 1);
        return {result[1], result[2]};
    }
};

// Time:  O(nlogn)
// Space: O(n)
// divide and conquer, merge sort, variant of closest pair
// reference: https://www.baeldung.com/cs/minimal-manhattan-distance
class Solution3 {
private:
    template<typename T>
    struct VectorHash {
        size_t operator()(const std::vector<T>& v) const {
            size_t seed = 0;
            for (const auto& i : v) {
                seed ^= std::hash<T>{}(i)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            }
            return seed;
        }
    };

public:
    vector<int> beautifulPair(vector<int>& nums1, vector<int>& nums2) {
        static const int INF = numeric_limits<int>::max();
        static const int MAX_NEIGHBOR_COUNT = 8;

        vector<vector<int>> points;
        for (int i = 0; i < size(nums1); ++i) {
            points.push_back({nums1[i], nums2[i]});
        }
        vector<int> result(3, INF);
        unordered_map<vector<int>, int, VectorHash<int>> lookup;
        for (int i = size(points) - 1; i >= 0; --i) {
            if (lookup.count(points[i])) {
                result = min(result, vector<int>{0, i, lookup[points[i]]});
            }
            lookup[points[i]] = i;
        }
        if (result[0] == 0) {
            return {result[1], result[2]};
        }
        vector<int> order(size(points));
        iota(begin(order), end(order), 0);
        sort(begin(order), end(order), [&](const auto& a, const auto& b) {
            return points[a][0] < points[b][0];
        });
        const auto& dist = [&](auto a, auto b) {
            if (a > b) {
                swap(a, b);
            }
            return vector<int>{abs(points[a][0] - points[b][0]) + abs(points[a][1] - points[b][1]), a, b};
        };

        const function<void (int, int)> merge_sort = [&](int left, int right) {
            if (left == right) {
                return;
            }
            const int mid = left + (right - left) / 2;
            const int x = points[order[mid]][0];  // added
            merge_sort(left, mid);
            merge_sort(mid + 1, right);
            vector<int> tmp;
            for (int l = left, r = mid + 1; l <= mid; ++l) {
                while (r <= right && points[order[r]][1] < points[order[l]][1]) {  // modified
                    tmp.emplace_back(order[r++]);
                }
                tmp.emplace_back(order[l]);
            }
            copy(cbegin(tmp), cend(tmp), begin(order) + left);

            // added below
            vector<int> stripe;
            for (int i = left; i <= right; ++i) {
                if (abs(points[order[i]][0] - x) <= result[0]) {
                    stripe.emplace_back(order[i]);
                }
            }
            for (int i = 0; i + 1 < size(stripe); ++i) {
                for (int j = i + 1; j < size(stripe); ++j) {
                    int x = stripe[i], y = stripe[j];
                    if (points[y][1] - points[x][1] > result[0]) {
                        break;
                    }
                    assert(j - (i + 1) + 1 <= MAX_NEIGHBOR_COUNT);
                    result = min(result, dist(x, y));
                }
            }      
        };
        
        merge_sort(0, size(points) - 1);
        return {result[1], result[2]};
    }
};

// Time:  O(nlogn)
// Space: O(n)
// segment tree
class Solution4 {
private:
    template<typename T>
    struct VectorHash {
        size_t operator()(const std::vector<T>& v) const {
            size_t seed = 0;
            for (const auto& i : v) {
                seed ^= std::hash<T>{}(i)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            }
            return seed;
        }
    };

public:
    vector<int> beautifulPair(vector<int>& nums1, vector<int>& nums2) {
        static const int INF = numeric_limits<int>::max();

        vector<vector<int>> points;
        for (int i = 0; i < size(nums1); ++i) {
            points.push_back({nums1[i], nums2[i]});
        }
        vector<int> result(3, INF);
        unordered_map<vector<int>, int, VectorHash<int>> lookup;
        for (int i = size(points) - 1; i >= 0; --i) {
            if (lookup.count(points[i])) {
                result = min(result, vector<int>{0, i, lookup[points[i]]});
            }
            lookup[points[i]] = i;
        }
        if (result[0] == 0) {
            return {result[1], result[2]};
        }
        vector<int> order(size(points));
        iota(begin(order), end(order), 0);
        sort(begin(order), end(order), [&](const auto& a, const auto& b) {
            return points[a][0] < points[b][0];
        });
        const auto& dist = [&](auto a, auto b) {
            if (a > b) {
                swap(a, b);
            }
            return vector<int>{abs(points[a][0] - points[b][0]) + abs(points[a][1] - points[b][1]), a, b};
        };

        unordered_set<int> y_set;
        for (const auto& p : points) {
            y_set.emplace(p[1]);
        }
        vector<int> sorted_y(cbegin(y_set), cend(y_set));
        sort(begin(sorted_y), end(sorted_y));
        unordered_map<int, int> y_to_idx;
        for (int i = 0; i < size(sorted_y); ++i) {  // coordinate compression
            y_to_idx[sorted_y[i]] = i;
        }
        SegmentTree st1(size(y_to_idx)), st2(size(y_to_idx));
        for (const auto& i : order) {
            int j = -st1.query(0, y_to_idx[points[i][1]] - 1)[1];  // min((xi-xj)+(yi-yj) for j in range(y_to_idx[points[i][1])) = (xi+yi)-max((xj+yj) for j in range(y_to_idx[points[i][1]))
            if (j != INF) {
                assert(points[j][1] < points[i][1]);
                result = min(result, dist(i, j));
            }
            st1.update(y_to_idx[points[i][1]], vector<int>{points[i][0] + points[i][1], -i});
            j = -st2.query(y_to_idx[points[i][1]], size(y_to_idx) - 1)[1];   // min((xi-xj)+(yj-yi) for j in range(y_to_idx[points[i][1], len(y_to_idx))) = (xi-yi)-max((xj-yj) for j in range(y_to_idx[points[i][1], len(y_to_idx))
            if (j != INF) {
                assert(points[j][1] >= points[i][1]);
                result = min(result, dist(i, j));
            }
            st2.update(y_to_idx[points[i][1]], vector<int>{points[i][0] - points[i][1], -i});
        }
        return {result[1], result[2]};
    }

private:
    class SegmentTree {
    private:
        static const int INF = numeric_limits<int>::max();
      
    public:
        explicit SegmentTree(int N)
          : tree(N > 1 ? 1 << (__lg(N - 1) + 2) : 2, vector<int>{-INF, -INF}),
            base(N > 1 ? 1 << (__lg(N - 1) + 1) : 1) {

        }

        void update(int i, const vector<int>& h) {
            int x = base + i;
            tree[x] = h;
            while (x > 1) {
                x /= 2;
                tree[x] = max(tree[x * 2], tree[x * 2 + 1]);
            }
        }

        vector<int> query(int L, int R) {
            vector<int> result = {-INF, -INF};
            if (L > R) {
                return result;
            }
            L += base;
            R += base;
            for (; L <= R; L /= 2, R /= 2) {
                if (L & 1) {
                    result = max(result, tree[L]);
                    ++L;
                }
                if ((R & 1) == 0) {
                    result = max(tree[R], result);
                    --R;
                }
            }
            return result;
        }

        vector<vector<int>> tree;
        int base;
    };
};

Beginner Explanation

What is Beautiful Pairs?

Beautiful Pairs (LeetCode #2613) is a Hard problem that primarily trains recursion.

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 random algorithms, merge sort, and segment 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: Random Algorithms, Divide and Conquer, Merge Sort, Segment Tree.

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

Interview Walkthrough

Interview approach for Beautiful Pairs

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 random algorithms, merge sort, and segment 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) on average) 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) on average time and O(n) space.

Pattern focus: random algorithms, merge sort, and segment tree

Use the pattern as a checklist:

random algorithms — confirm the invariant holds after each step
merge sort — confirm the invariant holds after each step
segment 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) on average
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 Beautiful Pairs

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for random algorithms, merge sort, and segment 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 random algorithms, merge sort, and segment 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: recursion.

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 Beautiful Pairs in a second language (cpp, python).
Drill 3–5 more problems tagged recursion.
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 Beautiful Pairs: show input structure (recursion), highlight the moving parts of the random algorithms, merge sort, and segment 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 random algorithms, merge sort, and segment 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

Beautiful Pairs (#2613) — Hard. Pattern: random algorithms, merge sort, and segment tree. Complexity: O(n) on average time / O(n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Beautiful Pairs?+

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

What pattern does Beautiful Pairs use?+

It primarily maps to random algorithms, merge sort, and segment tree, within the broader topic of recursion.

Is Beautiful Pairs 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/beautiful-pairs/

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