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 i
th job, and profit[i]
is the profit of the i
th job.
Now we have some workers. worker[i]
is the ability of the i
th 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.
// OJ: https://leetcode.com/problems/mostprofitassigningwork/
// 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;
}
};
Java

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; } return totalProfit; } 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]; } }

// OJ: https://leetcode.com/problems/mostprofitassigningwork/ // 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(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