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Problem solution · C++

Distinct Gate Paths to Lca

Distinct Gate Paths to Lca: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
103 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

Depth-first search

For Distinct Gate Paths to Lca, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 103 lines of C++ from the credited upstream file distinct-gate-paths-to-lca.cpp.
  • The implementation visibly relies on sequence storage.
  • 10 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeDistinct Gate Paths to Lca · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(nlogn + qlogn)// Space: O(nlogn) // dfs, binary liftingclass Solution {public:    int distinctPaths(int n, vector<int>& parent, vector<vector<int>>& gates, vector<vector<int>>& queries) {        static const int64_t MOD = 1e9 + 7;        enum { BB, BR, RB, RR };        using Mat = array<int64_t, 4>;         const auto& ceil_log2 = [&](uint32_t x) {            return bit_width(x - 1);        };         const auto& mult = [&](const auto& x, const auto& y) -> Mat {            return {                (x[BB] * y[BB] + x[BR] * y[RB]) % MOD,                (x[BB] * y[BR] + x[BR] * y[RR]) % MOD,                (x[RB] * y[BB] + x[RR] * y[RB]) % MOD,                (x[RB] * y[BR] + x[RR] * y[RR]) % MOD,            };        };         vector<int> depth(n);        vector<vector<int>> par(ceil_log2(n - 1) + 1, vector<int>(n));        const auto& lca = [&](int a, int b) {            if (depth[a] < depth[b]) {                swap(a, b);            }            const auto& d = depth[a] - depth[b];            for (int k = 0; k < size(par); ++k) {                if (d & (1 << k)) {                    a = par[k][a];                }            }            if (a == b) {                return a;            }            for (int k = size(par) - 1; k >= 0; --k) {                if (par[k][a] == par[k][b]) {                    continue;                }                a = par[k][a];                b = par[k][b];            }            return par[0][a];        };         vector<vector<Mat>> cnt(ceil_log2(n - 1) + 1, vector<Mat>(n));        const auto& count = [&](int u, int card, int t) -> int64_t {            if (u == t) {                return 1;            }            const auto& d = depth[u] - depth[t];            int64_t b = (card == 0) ? 1 : 0;            int64_t r = (card == 1) ? 1 : 0;            for (int k = 0; k < size(par); ++k) {                if (!(d & (1 << k))) {                    continue;                }                tie(b, r) = pair((b * cnt[k][u][BB] + r * cnt[k][u][RB]) % MOD, (b * cnt[k][u][BR] + r * cnt[k][u][RR]) % MOD);                u = par[k][u];            }            return (b + r) % MOD;        };         vector<vector<int>> adj(n);        for (int u = 0; u < n; ++u) {            if (parent[u] != -1) {                adj[parent[u]].emplace_back(u);            }        }        vector<int> stk = {0};        while (!empty(stk)) {            const auto u = stk.back(); stk.pop_back();            for (const auto& v : adj[u]) {                depth[v] = depth[u] + 1;                stk.emplace_back(v);            }        }        for (int u = 0; u < n; ++u) {            par[0][u] = parent[u];            cnt[0][u] = {gates[u][1], gates[u][2], gates[u][2], gates[u][0]};        }        for (int k = 1; k < size(par); ++k) {            for (int u = 0; u < n; ++u) {                if (par[k - 1][u] == -1) {                    continue;                }                par[k][u] = par[k - 1][par[k - 1][u]];                cnt[k][u] = mult(cnt[k - 1][u], cnt[k - 1][par[k - 1][u]]);            }        }        int result = 0;        for (const auto& q : queries) {            const auto& l = lca(q[0], q[2]);            result ^= (count(q[0], q[1], l) * count(q[2], q[3], l)) % MOD;        }        return result;    }};

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