How do you solve Most Profit Assigning Work on LeetCode?

Most Profit Assigning Work (LeetCode #826) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log m) time complexity and O(n) space complexity. Key insight: Workers can only take jobs they can handle, so sorting helps match them efficiently.

What is the time complexity of Most Profit Assigning Work?

The optimal solution for Most Profit Assigning Work runs in O(n log n + m log m) time and uses O(n) space. The time complexity is O(n log n + m log m) due to the sorting of jobs and workers. The space complexity is O(n) for storing the job pairs.

What is the brute force solution for Most Profit Assigning Work?

The brute force approach for Most Profit Assigning Work is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every worker against every job to see if they can complete it. If they can, we calculate the profit and keep track of the maximum profit found. The optimal Optimal Solution approach improves this to O(n log n + m log m).

What algorithm or data structure is used to solve Most Profit Assigning Work?

Most Profit Assigning Work uses the following concepts: Array, Two Pointers, Binary Search, Greedy, Sorting. The recommended patterns to study are: Sorting, Two Pointers.

What are the interview tips for Most Profit Assigning Work?

Always consider sorting when dealing with ranges or limits. Think about how to reduce redundant calculations — use pointers or flags. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Most Profit Assigning Work?

Not sorting the jobs and workers, which leads to inefficient checks. Failing to update the maximum profit correctly when multiple workers can do the same job.

#826

Most Profit Assigning Work

Medium

ArrayTwo PointersBinary SearchGreedySortingSortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log m)
Space	O(1)	O(n)

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Intuition

Time O(n log n + m log m)Space O(n)

In the optimal solution, we first pair jobs with their profits and sort them. Then, we sort the workers and use a two-pointer technique to efficiently find the maximum profit for each worker based on their ability.

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Algorithm

3 steps

1Step 1: Pair the difficulty and profit into a list of tuples and sort them based on difficulty.
2Step 2: Sort the workers array.
3Step 3: Use a pointer to track the maximum profit available for jobs that can be done by the current worker, and iterate through the workers to accumulate the total profit.

solution.py17 lines

1# Full working Python code
2
3def maxProfit(difficulty, profit, worker):
4    jobs = sorted(zip(difficulty, profit))
5    worker.sort()
6    total_profit = 0
7    max_profit = 0
8    j = 0
9    for w in worker:
10        while j < len(jobs) and jobs[j][0] <= w:
11            max_profit = max(max_profit, jobs[j][1])
12            j += 1
13        total_profit += max_profit
14    return total_profit
15
16# Example usage
17print(maxProfit([2,4,6,8,10], [10,20,30,40,50], [4,5,6,7]))

ℹ

Complexity note: The time complexity is O(n log n + m log m) due to the sorting of jobs and workers. The space complexity is O(n) for storing the job pairs.

1Workers can only take jobs they can handle, so sorting helps match them efficiently.
2Using a two-pointer technique allows us to keep track of the maximum profit without redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.