Cluster Process Scheduling Editorial

Overview

The problem asks us to count the number of ways to assign nProcesses to nIntervals such that no two adjacent intervals contain the same process. This is a classic combinatorics problem that can be solved by considering the number of choices available for each time slot.

For the first time slot, we have a free choice of any of the nProcesses. For every subsequent slot, we can choose any process except the one used in the immediately preceding slot. This ensures the "no consecutive" rule is satisfied.

Brute Force

A brute force approach would involve using recursion or backtracking to explore every possible valid sequence of processes. We would try all nProcesses for the first slot, and for each following slot, we would try all processes that are not equal to the previous one.

While this correctly counts the schedules, the number of valid schedules grows exponentially, and a search space of $O(P \cdot (P-1)^{I-1})$ is far too large for the given constraints where nIntervals can be $10^9$.

def countValidSchedules(nProcesses: int, nIntervals: int) -> int:
    MOD = 10**9 + 7
    
    def backtrack(current_interval, last_process):
        if current_interval == nIntervals:
            return 1
        
        count = 0
        for p in range(nProcesses):
            if p != last_process:
                count = (count + backtrack(current_interval + 1, p)) % MOD
        return count

    # Start with -1 as last_process to allow any process in the first slot
    return backtrack(0, -1)

Time complexity: O(nProcesses^nIntervals) — Each of the nIntervals can theoretically branch into nProcesses choices, leading to exponential growth. Space complexity: O(nIntervals) — The recursion stack depth is equal to the number of intervals.

Optimal / Best Approach

The optimal approach utilizes the fundamental counting principle. We can break down the decision-making process slot by slot:

1. First Slot: We have nProcesses choices.
2. Second Slot: We must pick a process different from the first. This leaves us with (nProcesses - 1) choices.
3. Third Slot: We must pick a process different from the second. Even though we could technically reuse the process from the first slot, the only restriction is the second slot. Thus, we again have (nProcesses - 1) choices.
4. $i$-th Slot: For any slot from $2$ to $n$, we always have (nProcesses - 1) choices.

The total number of ways is the product of these choices: $$Total = nProcesses \times (nProcesses - 1)^{nIntervals - 1}$$

Since $nIntervals$ is large, we use Modular Exponentiation (the pow function in Python) to calculate $(nProcesses - 1)^{nIntervals - 1} \pmod{10^9 + 7}$ in logarithmic time.

def countValidSchedules(nProcesses: int, nIntervals: int) -> int:
    if nIntervals == 0:
        return 0
    if nProcesses == 0:
        return 0
    if nProcesses == 1:
        # If there is 1 process, it can only fill 1 interval.
        # More than 1 interval forces a consecutive repeat.
        return 1 if nIntervals == 1 else 0
    
    MOD = 10**9 + 7
    
    # Formula: nProcesses * (nProcesses - 1)^(nIntervals - 1)
    # Use pow(base, exp, mod) for efficient calculation
    term1 = nProcesses % MOD
    term2 = pow(nProcesses - 1, nIntervals - 1, MOD)
    
    return (term1 * term2) % MOD

Time complexity: O(log nIntervals) — Modular exponentiation takes logarithmic time relative to the exponent. Space complexity: O(1) — We only store a few integer variables regardless of the input size.

Edge Cases

• nProcesses = 1: If there is more than one interval, it is impossible to avoid consecutive duplicates. The result is 1 if nIntervals == 1, else 0.
• nIntervals = 1: The answer is simply nProcesses, as any single process choice is valid.
• Large Inputs: The use of pow(a, b, m) is critical here to prevent integer overflow and to handle the $10^9$ constraints within the time limit.

Fast