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Problem solution · C++

Finish Time of Tasks I

Finish Time of Tasks I: 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
68 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 Finish Time of Tasks I, 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

  • 68 lines of C++ from the credited upstream file finish-time-of-tasks-i.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 6 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 codeFinish Time of Tasks I · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n)// Space: O(n) // iterative dfs, tree dpclass Solution {public:    long long finishTime(int n, vector<vector<int>>& edges, vector<int>& baseTime) {        static const auto& POS_INF = numeric_limits<int64_t>::max();        static const auto& NEG_INF = numeric_limits<int64_t>::min();         vector<vector<int>> adj(n);        const auto iter_dfs = [&]() {            vector<int64_t> dp(n);            vector<pair<int, int>> stk = {{1, 0}};            while (!empty(stk)) {                const auto [step, u] = stk.back(); stk.pop_back();                if (step == 1) {                    stk.emplace_back(2, u);                    for (const auto& v : adj[u]) {                        stk.emplace_back(1, v);                    }                } else if (step == 2) {                    auto mx = numeric_limits<int64_t>::min();                    auto mn = numeric_limits<int64_t>::max();                    for (const auto& v : adj[u]) {                        mx = max(mx, dp[v]);                        mn = min(mn, dp[v]);                    }                    dp[u] = (mx != NEG_INF ? (2 * mx - mn) : 0) + baseTime[u];                }            }            return dp[0];        };         for (const auto& e : edges) {            adj[e[0]].emplace_back(e[1]);        }        return iter_dfs();    }}; // Time:  O(n)// Space: O(n)// dfs, tree dpclass Solution2 {public:    long long finishTime(int n, vector<vector<int>>& edges, vector<int>& baseTime) {        static const auto& POS_INF = numeric_limits<int64_t>::max();        static const auto& NEG_INF = numeric_limits<int64_t>::min();         vector<vector<int>> adj(n);        const auto dfs = [&](this auto&& dfs, int u) -> int64_t {            auto mx = NEG_INF, mn = POS_INF;            for (const auto& v : adj[u]) {                const auto& ret = dfs(v);                mx = max(mx, ret);                mn = min(mn, ret);            }            return (mx != NEG_INF ? (2 * mx - mn) : 0) + baseTime[u];        };         for (const auto& e : edges) {            adj[e[0]].emplace_back(e[1]);        }        return dfs(0);    }};

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