# 1235.Maximum-Profit-in-Job-Scheduling

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/maximum-profit-in-job-scheduling/

## ้ข็ฎๆ่ฟฐ

``````We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTime, endTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

Example 1:
Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job.
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:
Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job.
Profit obtained 150 = 20 + 70 + 60.

Example 3:
Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

Constraints:
1 <= startTime.length == endTime.length == profit.length <= 5 * 104
1 <= startTime[i] < endTime[i] <= 109
1 <= profit[i] <= 104``````

## ไปฃ็ 

### Approach #1 DP

``````class Solution {
public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
int n = startTime.length;
int[][] jobs = new int[n][3];
for (int i = 0; i < n; i++) {
jobs[i][0] = startTime[i];
jobs[i][1] = endTime[i];
jobs[i][2] = profit[i];
}

Arrays.sort(jobs, (a, b) -> a[1] - b[1]);

int[] dp = new int[n];
dp[0] = jobs[0][2];

for (int i = 1; i < n; i++) {
dp[i] = Math.max(jobs[i][2], dp[i-1]);
for (int j = i - 1; j >= 0; j--) {
if (jobs[j][1] <= jobs[i][0]) {
dp[i] = Math.max(dp[i], jobs[i][2] + dp[j]);
break; // ไธๅ ่ฟไธชไผ่ถๆถ
}
}
}

int ans = Integer.MIN_VALUE;
for (int val: dp) {
ans = Math.max(ans, val);
}
return ans;
}
}``````

### Approach #2 TreeMap

Time: O(nlogn) && Space: O(n)

``````class Solution {
public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
int n = startTime.length;
int[][] jobs = new int[n][3];
for (int i = 0; i < n; i++) {
jobs[i][0] = startTime[i];
jobs[i][1] = endTime[i];
jobs[i][2] = profit[i];
}

Arrays.sort(jobs, (a, b) -> a[1] - b[1]);
TreeMap<Integer, Integer> map = new TreeMap();
int ans = 0;

for (int i = 0; i < n; i++) {
Integer start = jobs[i][0];
Integer prev = map.floorKey(start);
if (prev == null) {
ans = Math.max(ans, jobs[i][2]);
} else {
int preSum = map.get(prev);
ans = Math.max(ans, preSum + jobs[i][2]);
}
map.put(jobs[i][1], ans);
}

return ans;
}
}``````

### Approach #3 DFS

``````class Solution {
public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
int n = startTime.length;
int[][] jobs = new int[n][3];
for (int i = 0; i < n; i++) {
jobs[i][0] = startTime[i];
jobs[i][1] = endTime[i];
jobs[i][2] = profit[i];
}

Arrays.sort(jobs, (a, b) -> a[0] - b[0]);

Map<Integer, Integer> memo = new HashMap();

return dfs(jobs, 0, memo);
}

private int dfs( int[][] jobs, int index, Map<Integer, Integer> memo) {
int n = jobs.length;
if (n == index)   return 0;
if (memo.containsKey(index))    return memo.get(index);

/// replaced by binary search  => nlogn
for (int next = index + 1; next < n; next++) {
if (jobs[next][0] >= jobs[index][1]) {
int include = jobs[index][2] + dfs(jobs, next, memo);
int exclude = dfs(jobs, index + 1, memo);
int max = Math.max(include, exclude);
memo.put(index, max);
return max;
}
}

int include = jobs[index][2];
int exclude = dfs(jobs, index + 1, memo);
int max = Math.max(include, exclude);
memo.put(index, max);
return max;
}
}``````

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