LeetCode #2895 — MEDIUM

Minimum Processing Time

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once.

You are given an array processorTime representing the time each processor becomes available and an array tasks representing how long each task takes to complete. Return the minimum time needed to complete all tasks.

Example 1:

Input: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]

Output: 16

Explanation:

Assign the tasks at indices 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indices 0, 1, 2, 3 to the second processor which becomes available at time = 10.

The time taken by the first processor to finish the execution of all tasks is max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.

The time taken by the second processor to finish the execution of all tasks is max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.

Example 2:

Input: processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]

Output: 23

Explanation:

Assign the tasks at indices 1, 4, 5, 6 to the first processor and the others to the second processor.

The time taken by the first processor to finish the execution of all tasks is max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.

The time taken by the second processor to finish the execution of all tasks is max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.

Constraints:

1 <= n == processorTime.length <= 25000
1 <= tasks.length <= 10⁵
0 <= processorTime[i] <= 10⁹
1 <= tasks[i] <= 10⁹
tasks.length == 4 * n

Step 01

Brute Force Baseline

Problem summary: You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once. You are given an array processorTime representing the time each processor becomes available and an array tasks representing how long each task takes to complete. Return the minimum time needed to complete all tasks.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[8,10]
[2,2,3,1,8,7,4,5]

Example 2

[10,20]
[2,3,1,2,5,8,4,3]

Step 02

Core Insight

What unlocks the optimal approach

It’s optimal to make the processor with earlier process time run 4 longer tasks.****
The largest <code>processTime[i] + tasks[j]</code> (when matched) is the answer.

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

Upper-end input sizes

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 #2895: Minimum Processing Time
class Solution {
    public int minProcessingTime(List<Integer> processorTime, List<Integer> tasks) {
        processorTime.sort((a, b) -> a - b);
        tasks.sort((a, b) -> a - b);
        int ans = 0, i = tasks.size() - 1;
        for (int t : processorTime) {
            ans = Math.max(ans, t + tasks.get(i));
            i -= 4;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2895: Minimum Processing Time
func minProcessingTime(processorTime []int, tasks []int) (ans int) {
	sort.Ints(processorTime)
	sort.Ints(tasks)
	i := len(tasks) - 1
	for _, t := range processorTime {
		ans = max(ans, t+tasks[i])
		i -= 4
	}
	return
}

# Accepted solution for LeetCode #2895: Minimum Processing Time
class Solution:
    def minProcessingTime(self, processorTime: List[int], tasks: List[int]) -> int:
        processorTime.sort()
        tasks.sort()
        ans = 0
        i = len(tasks) - 1
        for t in processorTime:
            ans = max(ans, t + tasks[i])
            i -= 4
        return ans

// Accepted solution for LeetCode #2895: Minimum Processing Time
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2895: Minimum Processing Time
// class Solution {
//     public int minProcessingTime(List<Integer> processorTime, List<Integer> tasks) {
//         processorTime.sort((a, b) -> a - b);
//         tasks.sort((a, b) -> a - b);
//         int ans = 0, i = tasks.size() - 1;
//         for (int t : processorTime) {
//             ans = Math.max(ans, t + tasks.get(i));
//             i -= 4;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2895: Minimum Processing Time
function minProcessingTime(processorTime: number[], tasks: number[]): number {
    processorTime.sort((a, b) => a - b);
    tasks.sort((a, b) => a - b);
    let [ans, i] = [0, tasks.length - 1];
    for (const t of processorTime) {
        ans = Math.max(ans, t + tasks[i]);
        i -= 4;
    }
    return ans;
}

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 × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.