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Problem solution · C++

Find Weighted Median Node in Tree

Find Weighted Median Node in Tree: a C++ solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Disjoint set union
Source
Kamyu LeetCode Solutions
Length
271 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Find Weighted Median Node in Tree, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 271 lines of C++ from the credited upstream file find-weighted-median-node-in-tree.cpp.
  • The implementation visibly relies on sequence storage.
  • 21 loop blocks detected, together with recursive traversal.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Weighted Median Node in Tree · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n + qlogh)// Space: O(n + q) class UnionFind {public:    UnionFind(int n)        : set_(n)        , rank_(n) {         iota(begin(set_), end(set_), 0);    }     int find_set(int x) {        vector<int> stk;        while (set_[x] != x) {  // path compression            stk.emplace_back(x);            x = set_[x];        }        while (!empty(stk)) {            const int y = stk.back(); stk.pop_back();            set_[y] = x;        }        return x;    }     bool union_set(int x, int y) {        x = find_set(x), y = find_set(y);        if (x == y) {            return false;        }        if (rank_[x] > rank_[y]) {            swap(x, y);        }        set_[x] = y;  // Union by rank.        if (rank_[x] == rank_[y]) {            ++rank_[y];        }        return true;    } private:    vector<int> set_;    vector<int> rank_;}; int binary_search(int left, int right, const auto& check) {    while (left <= right) {        const auto& mid = left + (right - left) / 2;        if (check(mid)) {            right = mid - 1;        } else {            left = mid + 1;        }    }    return left;}; // iterative dfs, Tarjan's Offline LCA Algorithm, binary search, prefix sumclass Solution {public:    vector<int> findMedian(int n, vector<vector<int>>& edges, vector<vector<int>>& queries) {        vector<vector<pair<int, int>>> adj(size(edges) + 1);        for (const auto& e : edges) {            adj[e[0]].emplace_back(e[1], e[2]);            adj[e[1]].emplace_back(e[0], e[2]);        }        const auto& iter_dfs = [&]() {            vector<bool> lookup(size(adj));            vector<vector<int>> lookup2(size(adj));            for (int i = 0; i < size(queries); ++i) {                for (const auto& x : queries[i]) {                    lookup2[x].emplace_back(i);                }            }            UnionFind uf(size(adj));            vector<int> ancestor(size(adj));            iota(begin(ancestor), end(ancestor), 0);            vector<int64_t> dist(size(adj));            vector<int> depth(size(adj));            vector<int> lca(size(queries));            vector<int64_t> result(size(queries));            vector<tuple<int, int, int, int>> stk = {{1, 0, -1, -1}};            while (!empty(stk)) {                const auto [step, u, p, i] = stk.back(); stk.pop_back();                if (step == 1) {                    for (const auto& i : lookup2[u]) {                        if (queries[i][0] == queries[i][1]) {                            lca[i] = u;                            continue;                        }                        result[i] += dist[u];                        for (const auto& x : queries[i]) {                            if (lookup[x]) {                                lca[i] = ancestor[uf.find_set(x)];                                result[i] -= 2 * dist[lca[i]];                            }                        }                    }                    lookup[u] = true;                    stk.emplace_back(2, u, -1, 0);                } else if (step == 2) {                    if (i == size(adj[u])) {                        continue;                    }                    const auto& [v, w] = adj[u][i];                    stk.emplace_back(2, u, -1, i + 1);                    if (lookup[v]) {                        continue;                    }                    dist[v] = dist[u] + w;                    depth[v] = depth[u] + 1;                    stk.emplace_back(3, v, u, -1);                    stk.emplace_back(1, v, -1, -1);                } else if (step == 3) {                    uf.union_set(u, p);                    ancestor[uf.find_set(p)] = p;                }            }            return tuple(result, lca, dist, depth);        };         const auto& [result, lca, dist, depth] = iter_dfs();        const auto& iter_dfs2 = [&]() {            vector<vector<pair<int, int>>> lookup3(size(adj));            for (int i = 0; i < size(queries); ++i) {                const int u = queries[i][0], v = queries[i][1];                if (2 * (dist[u] - dist[lca[i]]) >= result[i]) {                    lookup3[u].emplace_back(i, 0);                } else {                    lookup3[v].emplace_back(i, 1);                }            }            vector<int> result2(size(queries));            vector<int> path;            vector<tuple<int, int, int>> stk = {{1, 0, -1}};            while (!empty(stk)) {                const auto [step, u, i] = stk.back(); stk.pop_back();                if (step == 1) {                    path.emplace_back(u);                    for (const auto& [i, t] : lookup3[u]) {                        const auto& d = depth[u] - depth[lca[i]];                        if (t == 0) {                            const auto& j = binary_search(0, d, [&](const auto& x) {                                return 2 * (dist[u] - dist[path[size(path) - (x + 1)]]) >= result[i];                            });                            result2[i] = path[size(path) - (j + 1)];                        } else {                            const auto& l = dist[queries[i][0]] - dist[lca[i]];                            const auto& j = binary_search(0, d - 1, [&](const auto& x) {                                return 2 * (l + (dist[path[size(path) - ((d - 1) + 1) + x]] - dist[lca[i]])) >= result[i];                            });                            result2[i] = path[size(path) - ((d - 1) + 1) + j];                        }                    }                    stk.emplace_back(3, u, -1);                    stk.emplace_back(2, u, 0);                } else if (step == 2) {                    if (i == size(adj[u])) {                        continue;                    }                    const auto& [v, w] = adj[u][i];                    stk.emplace_back(2, u, i + 1);                    if (size(path) >= 2 && path[size(path) - 2] == v) {                        continue;                    }                    stk.emplace_back(1, v, -1);                } else if (step == 3) {                    path.pop_back();                }            }            return result2;        };            return iter_dfs2();    }}; // Time:  O(n + qlogh)// Space: O(n + q)// dfs, Tarjan's Offline LCA Algorithm, binary search, prefix sumclass Solution2 {public:    vector<int> findMedian(int n, vector<vector<int>>& edges, vector<vector<int>>& queries) {        vector<vector<pair<int, int>>> adj(size(edges) + 1);        for (const auto& e : edges) {            adj[e[0]].emplace_back(e[1], e[2]);            adj[e[1]].emplace_back(e[0], e[2]);        }        vector<bool> lookup(size(adj));        vector<vector<int>> lookup2(size(adj));        for (int i = 0; i < size(queries); ++i) {            for (const auto& x : queries[i]) {                lookup2[x].emplace_back(i);            }        }        UnionFind uf(size(adj));        vector<int> ancestor(size(adj));        iota(begin(ancestor), end(ancestor), 0);        vector<int64_t> dist(size(adj));        vector<int> depth(size(adj));        vector<int> lca(size(queries));        vector<int64_t> result(size(queries));        const function<void (int)> dfs = [&](int u) {            for (const auto& i : lookup2[u]) {                if (queries[i][0] == queries[i][1]) {                    lca[i] = u;                    continue;                }                result[i] += dist[u];                for (const auto& x : queries[i]) {                    if (lookup[x]) {                        lca[i] = ancestor[uf.find_set(x)];                        result[i] -= 2 * dist[lca[i]];                    }                }            }            lookup[u] = true;            for (const auto& [v, w] : adj[u]) {                if (lookup[v]) {                    continue;                }                dist[v] = dist[u] + w;                depth[v] = depth[u] + 1;                dfs(v);                uf.union_set(v, u);                ancestor[uf.find_set(u)] = u;            }        };         dfs(0);        vector<int> result2(size(queries));        vector<vector<pair<int, int>>> lookup3(size(adj));        for (int i = 0; i < size(queries); ++i) {            const int u = queries[i][0], v = queries[i][1];            if (2 * (dist[u] - dist[lca[i]]) >= result[i]) {                lookup3[u].emplace_back(i, 0);            } else {                lookup3[v].emplace_back(i, 1);            }        }        vector<int> path;        const function<void (int)> dfs2 = [&](int u) {            path.emplace_back(u);            for (const auto& [i, t] : lookup3[u]) {                const auto& d = depth[u] - depth[lca[i]];                if (t == 0) {                    const auto& j = binary_search(0, d, [&](const auto& x) {                        return 2 * (dist[u] - dist[path[size(path) - (x + 1)]]) >= result[i];                    });                    result2[i] = path[size(path) - (j + 1)];                } else {                    const auto& l = dist[queries[i][0]] - dist[lca[i]];                    const auto& j = binary_search(0, d - 1, [&](const auto& x) {                        return 2 * (l + (dist[path[size(path) - ((d - 1) + 1) + x]] - dist[lca[i]])) >= result[i];                    });                    result2[i] = path[size(path) - ((d - 1) + 1) + j];                }            }            for (const auto& [v, w] : adj[u]) {                if (size(path) >= 2 && path[size(path) - 2] == v) {                    continue;                }                dfs2(v);            }            path.pop_back();        };            dfs2(0);        return result2;    }};

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