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 <= 100001 <= worker.length <= 10000difficulty[i], profit[i], worker[i]are in range[1, 10^5]
Companies:
Nutanix
Related Topics:
Two Pointers
Solution 1.
// 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;
}
};
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];
}
}