CPMagical CP Library

Problem solution · C++

Parallel Courses II

Parallel Courses II: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
40 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

Dynamic programming

For Parallel Courses II, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 40 lines of C++ from the credited upstream file 1494.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 4 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeParallel Courses II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minNumberOfSemesters(int n, vector<vector<int>>& relations, int k) {    // dp[i] := the minimum number of semesters to take the courses, where i is    // the bitmask of the taken courses    vector<int> dp(1 << n, n);    // prereq[i] := the bitmask of all the dependencies of the i-th course    vector<int> prereq(n);     for (const vector<int>& relation : relations) {      const int prevCourse = relation[0] - 1;      const int nextCourse = relation[1] - 1;      prereq[nextCourse] |= 1 << prevCourse;    }     dp[0] = 0;  // Don't need time to finish 0 course.     for (int i = 0; i < dp.size(); ++i) {      // the bitmask of all the courses can be taken      int coursesCanBeTaken = 0;      // Can take the j-th course if i contains all of j's prerequisites.      for (int j = 0; j < n; ++j)        if ((i & prereq[j]) == prereq[j])          coursesCanBeTaken |= 1 << j;      // Don't take any course which is already taken.      // (i represents set of courses that are already taken)      coursesCanBeTaken &= ~i;      // Enumerate every bitmask subset of `coursesCanBeTaken`.      for (unsigned s = coursesCanBeTaken; s > 0;           s = (s - 1) & coursesCanBeTaken)        if (popcount(s) <= k)          // Any combination of courses (if <= k) can be taken now.          // i | s := combining courses taken with courses can be taken.          dp[i | s] = min(dp[i | s], dp[i] + 1);    }     return dp.back();  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗