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Formatted question description: https://leetcode.ca/all/826.html

# 826. Most Profit Assigning Work (Medium)

We have jobs: difficulty[i] is the difficulty of the ith job, and profit[i] is the profit of the ith job.

Now we have some workers. worker[i] is the ability of the ith worker, which means that this worker can only complete a job with difficulty at most worker[i]

Every worker can be assigned at most one job, but one job can be completed multiple times.

For example, if 3 people attempt the same job that pays $1, then the total profit will be$3.  If a worker cannot complete any job, his profit is \$0.

What is the most profit we can make?

Example 1:

Input: difficulty = [2,4,6,8,10], profit = [10,20,30,40,50], worker = [4,5,6,7]
Output: 100
Explanation: Workers are assigned jobs of difficulty [4,4,6,6] and they get profit of [20,20,30,30] seperately.

Notes:

• 1 <= difficulty.length = profit.length <= 10000
• 1 <= worker.length <= 10000
• difficulty[i], profit[i], worker[i]  are in range [1, 10^5]

Companies:
Nutanix

Related Topics:
Two Pointers

## Solution 1.

• class Solution {
public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
int jobs = difficulty.length;
int[][] difficultyProfit = new int[jobs][2];
for (int i = 0; i < jobs; i++) {
difficultyProfit[i][0] = difficulty[i];
difficultyProfit[i][1] = profit[i];
}
Arrays.sort(difficultyProfit, new Comparator<int[]>() {
public int compare(int[] difficultyProfit1, int[] difficultyProfit2) {
if (difficultyProfit1[0] != difficultyProfit2[0])
return difficultyProfit1[0] - difficultyProfit2[0];
else
return difficultyProfit2[1] - difficultyProfit1[1];
}
});
for (int i = 1; i < jobs; i++)
difficultyProfit[i][1] = Math.max(difficultyProfit[i - 1][1], difficultyProfit[i][1]);
int totalProfit = 0;
for (int ability : worker) {
int maxProfit = binarySearch(difficultyProfit, ability);
totalProfit += maxProfit;
}
}

public int binarySearch(int[][] difficultyProfit, int ability) {
int low = 0, high = difficultyProfit.length - 1;
int index = -1;
while (low <= high) {
int mid = (high - low) / 2 + low;
int difficulty = difficultyProfit[mid][0];
if (difficulty == ability) {
index = mid;
break;
} else if (difficulty > ability)
high = mid - 1;
else
low = mid + 1;
}
if (index < 0)
index = low - 1;
if (index < 0)
return 0;
else
return difficultyProfit[index][1];
}
}

############

class Solution {
public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
int n = difficulty.length;
List<int[]> job = new ArrayList<>();
for (int i = 0; i < n; ++i) {
}
job.sort(Comparator.comparing(a -> a[0]));
Arrays.sort(worker);
int res = 0;
int i = 0, t = 0;
for (int w : worker) {
while (i < n && job.get(i)[0] <= w) {
t = Math.max(t, job.get(i++)[1]);
}
res += t;
}
return res;
}
}

• // OJ: https://leetcode.com/problems/most-profit-assigning-work/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int maxProfitAssignment(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
map<int, int, greater<int>> diffToProfit;
int N = difficulty.size(), maxProfit = 0, ans = 0;
for (int i = 0; i < N; ++i) {
diffToProfit[difficulty[i]] = max(diffToProfit[difficulty[i]], profit[i]);
}
for (auto it = diffToProfit.rbegin(); it != diffToProfit.rend(); ++it) {
it->second = maxProfit = max(maxProfit, it->second);
}
for (int w : worker) {
auto it = diffToProfit.lower_bound(w);
if (it != diffToProfit.end()) ans += it->second;
}
return ans;
}
};

• class Solution:
def maxProfitAssignment(
self, difficulty: List[int], profit: List[int], worker: List[int]
) -> int:
n = len(difficulty)
job = [(difficulty[i], profit[i]) for i in range(n)]
job.sort(key=lambda x: x[0])
worker.sort()
i = t = res = 0
for w in worker:
while i < n and job[i][0] <= w:
t = max(t, job[i][1])
i += 1
res += t
return res

############

class Solution(object):
def maxProfitAssignment(self, difficulty, profit, worker):
"""
:type difficulty: List[int]
:type profit: List[int]
:type worker: List[int]
:rtype: int
"""
jobs = sorted([a, b] for a, b in zip(difficulty, profit))
curMax, i = 0, 0
res = 0
for w in sorted(worker):
while i < len(jobs) and w >= jobs[i][0]:
curMax = max(curMax, jobs[i][1])
i += 1
res += curMax
return res

• func maxProfitAssignment(difficulty []int, profit []int, worker []int) int {
var job [][2]int
for i := range difficulty {
job = append(job, [2]int{difficulty[i], profit[i]})
}

sort.SliceStable(job, func(i, j int) bool { return job[i][0] <= job[j][0] })
sort.Ints(worker)
i, t, n, res := 0, 0, len(difficulty), 0
for _, w := range worker {
for i < n && job[i][0] <= w {
t = max(t, job[i][1])
i++
}
res += t
}
return res
}

func max(a, b int) int {
if a > b {
return a
}
return b
}