826 Most Profit Assigning Work

The problem comes off as pretty difficult at the start but the idea is very straight-forward, we just generate a lookup table for all difficulties.

We do this because the abilities of the workers can be anything from 0 to max(difficulty) where difficulty is the array given to us containing all the difficulty to profit mappings.

Since it’s told in the description that if any ability of a worker does not have a corresponding difficulty in the input, then to consider the smallest closest difficulty.

For eg.

Difficulty	Profit
3	20
5	35
8	45

Then a worker with ability of 3 will bring a profit of 20, and a worker with ability of 4 will also bring a profit of 20.

A worker with ability of 2 will bring 0 profit.

Code:

Python3

def maxProfitAssignment(self, difficulty: List[int], profit: List[int], worker: List[int]) -> int:
    maxDiff = max(difficulty)
 
    profitLookup = [0] * (maxDiff + 1)
 
    for i in range(len(difficulty)):
        profitLookup[difficulty[i]] = max(profitLookup[difficulty[i]], profit[i])
 
    for i in range(1, maxDiff + 1):
        profitLookup[i] = max(profitLookup[i], profitLookup[i-1])
 
    maxProfit = 0
    for ability in worker:
        maxProfit += profitLookup[min(maxDiff, ability)]
 
    return maxProfit

Big O Analysis

• Runtime

  The runtime complexity here is $O(maxDiff)$ where maxDiff = max(difficulty)

• Memory The memory usage is $O(N)$ since we use the profitLookup array to store cumulative profits for all difficulties up until maxDiff.

— A

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