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Problem solution · C++

Minimum Time to Complete All Tasks

Minimum Time to Complete All Tasks: a C++ solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
31 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

Sorting and greedy selection

For Minimum Time to Complete All Tasks, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 31 lines of C++ from the credited upstream file 2589.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Time to Complete All Tasks · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int findMinimumTime(vector<vector<int>>& tasks) {    constexpr int kMax = 2000;    vector<bool> running(kMax + 1);     // Sort tasks by end.    sort(        tasks.begin(), tasks.end(),        [](const vector<int>& a, const vector<int>& b) { return a[1] < b[1]; });     for (const vector<int>& task : tasks) {      const int start = task[0];      const int end = task[1];      const int duration = task[2];      int neededDuration = duration - count(running.begin() + start,                                            running.begin() + end + 1, true);      // Greedily run the task as late as possible so that later tasks can run      // simultaneously.      for (int i = end; neededDuration > 0; --i) {        if (!running[i]) {          running[i] = true;          --neededDuration;        }      }    }     return ranges::count(running, true);  }};

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