LeetCode #1494 — HARD

Parallel Courses II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given an array relations where relations[i] = [prevCourse_i, nextCourse_i], representing a prerequisite relationship between course prevCourse_i and course nextCourse_i: course prevCourse_i has to be taken before course nextCourse_i. Also, you are given the integer k.

In one semester, you can take at most k courses as long as you have taken all the prerequisites in the previous semesters for the courses you are taking.

Return the minimum number of semesters needed to take all courses. The testcases will be generated such that it is possible to take every course.

Example 1:

Input: n = 4, relations = [[2,1],[3,1],[1,4]], k = 2
Output: 3
Explanation: The figure above represents the given graph.
In the first semester, you can take courses 2 and 3.
In the second semester, you can take course 1.
In the third semester, you can take course 4.

Example 2:

Input: n = 5, relations = [[2,1],[3,1],[4,1],[1,5]], k = 2
Output: 4
Explanation: The figure above represents the given graph.
In the first semester, you can only take courses 2 and 3 since you cannot take more than two per semester.
In the second semester, you can take course 4.
In the third semester, you can take course 1.
In the fourth semester, you can take course 5.

Constraints:

1 <= n <= 15
1 <= k <= n
0 <= relations.length <= n * (n-1) / 2
relations[i].length == 2
1 <= prevCourse_i, nextCourse_i <= n
prevCourse_i != nextCourse_i
All the pairs [prevCourse_i, nextCourse_i] are unique.
The given graph is a directed acyclic graph.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given an array relations where relations[i] = [prevCoursei, nextCoursei], representing a prerequisite relationship between course prevCoursei and course nextCoursei: course prevCoursei has to be taken before course nextCoursei. Also, you are given the integer k. In one semester, you can take at most k courses as long as you have taken all the prerequisites in the previous semesters for the courses you are taking. Return the minimum number of semesters needed to take all courses. The testcases will be generated such that it is possible to take every course.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Bit Manipulation

Example 1

4
[[2,1],[3,1],[1,4]]
2

Example 2

5
[[2,1],[3,1],[4,1],[1,5]]
2

Core Insight

What unlocks the optimal approach

Use backtracking with states (bitmask, degrees) where bitmask represents the set of courses, if the ith bit is 1 then the ith course was taken, otherwise, you can take the ith course. Degrees represent the degree for each course (nodes in the graph).
Note that you can only take nodes (courses) with degree = 0 and it is optimal at every step in the backtracking take the maximum number of courses limited by k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1494: Parallel Courses II
class Solution {
    public int minNumberOfSemesters(int n, int[][] relations, int k) {
        int[] d = new int[n + 1];
        for (var e : relations) {
            d[e[1]] |= 1 << e[0];
        }
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0});
        Set<Integer> vis = new HashSet<>();
        vis.add(0);
        while (!q.isEmpty()) {
            var p = q.pollFirst();
            int cur = p[0], t = p[1];
            if (cur == (1 << (n + 1)) - 2) {
                return t;
            }
            int nxt = 0;
            for (int i = 1; i <= n; ++i) {
                if ((cur & d[i]) == d[i]) {
                    nxt |= 1 << i;
                }
            }
            nxt ^= cur;
            if (Integer.bitCount(nxt) <= k) {
                if (vis.add(nxt | cur)) {
                    q.offer(new int[] {nxt | cur, t + 1});
                }
            } else {
                int x = nxt;
                while (nxt > 0) {
                    if (Integer.bitCount(nxt) == k && vis.add(nxt | cur)) {
                        q.offer(new int[] {nxt | cur, t + 1});
                    }
                    nxt = (nxt - 1) & x;
                }
            }
        }
        return 0;
    }
}

// Accepted solution for LeetCode #1494: Parallel Courses II
func minNumberOfSemesters(n int, relations [][]int, k int) int {
	d := make([]int, n+1)
	for _, e := range relations {
		d[e[1]] |= 1 << e[0]
	}
	type pair struct{ v, t int }
	q := []pair{pair{0, 0}}
	vis := map[int]bool{0: true}
	for len(q) > 0 {
		p := q[0]
		q = q[1:]
		cur, t := p.v, p.t
		if cur == (1<<(n+1))-2 {
			return t
		}
		nxt := 0
		for i := 1; i <= n; i++ {
			if (cur & d[i]) == d[i] {
				nxt |= 1 << i
			}
		}
		nxt ^= cur
		if bits.OnesCount(uint(nxt)) <= k {
			if !vis[nxt|cur] {
				vis[nxt|cur] = true
				q = append(q, pair{nxt | cur, t + 1})
			}
		} else {
			x := nxt
			for nxt > 0 {
				if bits.OnesCount(uint(nxt)) == k && !vis[nxt|cur] {
					vis[nxt|cur] = true
					q = append(q, pair{nxt | cur, t + 1})
				}
				nxt = (nxt - 1) & x
			}
		}
	}
	return 0
}

# Accepted solution for LeetCode #1494: Parallel Courses II
class Solution:
    def minNumberOfSemesters(self, n: int, relations: List[List[int]], k: int) -> int:
        d = [0] * (n + 1)
        for x, y in relations:
            d[y] |= 1 << x
        q = deque([(0, 0)])
        vis = {0}
        while q:
            cur, t = q.popleft()
            if cur == (1 << (n + 1)) - 2:
                return t
            nxt = 0
            for i in range(1, n + 1):
                if (cur & d[i]) == d[i]:
                    nxt |= 1 << i
            nxt ^= cur
            if nxt.bit_count() <= k:
                if (nxt | cur) not in vis:
                    vis.add(nxt | cur)
                    q.append((nxt | cur, t + 1))
            else:
                x = nxt
                while nxt:
                    if nxt.bit_count() == k and (nxt | cur) not in vis:
                        vis.add(nxt | cur)
                        q.append((nxt | cur, t + 1))
                    nxt = (nxt - 1) & x

// Accepted solution for LeetCode #1494: Parallel Courses II
struct Solution;

struct Solver {
    n: usize,
    k: usize,
    pre: Vec<u32>,
    memo: Vec<i32>,
}

impl Solver {
    fn new(n: usize, k: usize, pre: Vec<u32>) -> Self {
        let mut memo: Vec<i32> = vec![-1; (1 << n) as usize];
        memo[0] = 0;
        Solver { n, k, pre, memo }
    }

    fn dfs(&self, k: usize, cur: u32, available: u32, groups: &mut Vec<u32>) {
        if k == 0 {
            groups.push(cur);
        }
        if available != 0 && k > 0 {
            let bit = 1 << available.trailing_zeros();
            self.dfs(k - 1, cur | bit, available & !bit, groups);
            self.dfs(k, cur, available & !bit, groups);
        }
    }

    fn dp(&mut self, left: u32) -> i32 {
        if self.memo[left as usize] != -1 {
            self.memo[left as usize]
        } else {
            let mut available: u32 = 0;
            for i in 0..self.n {
                if left & 1 << i != 0 && self.pre[i] & left == 0 {
                    available |= 1 << i;
                }
            }
            let res = if available.count_ones() <= self.k as u32 {
                1 + self.dp(left & !available)
            } else {
                let mut groups: Vec<u32> = vec![];
                self.dfs(self.k, 0, available, &mut groups);
                groups
                    .into_iter()
                    .map(|g| 1 + self.dp(left & !g))
                    .min()
                    .unwrap()
            };
            self.memo[left as usize] = res;
            res
        }
    }
}

impl Solution {
    fn min_number_of_semesters(n: i32, dependencies: Vec<Vec<i32>>, k: i32) -> i32 {
        let n = n as usize;
        let k = k as usize;
        let mut pre = vec![0; n];
        for dependency in dependencies {
            let x = dependency[0] as usize - 1;
            let y = dependency[1] as usize - 1;
            pre[y] |= 1 << x;
        }
        let mut solver = Solver::new(n, k, pre);
        solver.dp((1 << n) - 1)
    }
}

#[test]
fn test() {
    let n = 4;
    let dependencies = vec_vec_i32![[2, 1], [3, 1], [1, 4]];
    let k = 2;
    let res = 3;
    assert_eq!(Solution::min_number_of_semesters(n, dependencies, k), res);
    let n = 5;
    let dependencies = vec_vec_i32![[2, 1], [3, 1], [4, 1], [1, 5]];
    let k = 2;
    let res = 4;
    assert_eq!(Solution::min_number_of_semesters(n, dependencies, k), res);
    let n = 11;
    let dependencies = vec_vec_i32![];
    let k = 2;
    let res = 6;
    assert_eq!(Solution::min_number_of_semesters(n, dependencies, k), res);
}

// Accepted solution for LeetCode #1494: Parallel Courses II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1494: Parallel Courses II
// class Solution {
//     public int minNumberOfSemesters(int n, int[][] relations, int k) {
//         int[] d = new int[n + 1];
//         for (var e : relations) {
//             d[e[1]] |= 1 << e[0];
//         }
//         Deque<int[]> q = new ArrayDeque<>();
//         q.offer(new int[] {0, 0});
//         Set<Integer> vis = new HashSet<>();
//         vis.add(0);
//         while (!q.isEmpty()) {
//             var p = q.pollFirst();
//             int cur = p[0], t = p[1];
//             if (cur == (1 << (n + 1)) - 2) {
//                 return t;
//             }
//             int nxt = 0;
//             for (int i = 1; i <= n; ++i) {
//                 if ((cur & d[i]) == d[i]) {
//                     nxt |= 1 << i;
//                 }
//             }
//             nxt ^= cur;
//             if (Integer.bitCount(nxt) <= k) {
//                 if (vis.add(nxt | cur)) {
//                     q.offer(new int[] {nxt | cur, t + 1});
//                 }
//             } else {
//                 int x = nxt;
//                 while (nxt > 0) {
//                     if (Integer.bitCount(nxt) == k && vis.add(nxt | cur)) {
//                         q.offer(new int[] {nxt | cur, t + 1});
//                     }
//                     nxt = (nxt - 1) & x;
//                 }
//             }
//         }
//         return 0;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.