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
.
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.
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.
Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4] Output: 6
1 <= startTime.length == endTime.length == profit.length <= 5 * 104
1 <= startTime[i] < endTime[i] <= 109
1 <= profit[i] <= 104
impl Solution {
pub fn job_scheduling(start_time: Vec<i32>, end_time: Vec<i32>, profit: Vec<i32>) -> i32 {
let mut jobs = (0..profit.len())
.map(|i| (end_time[i], start_time[i], profit[i]))
.collect::<Vec<_>>();
let mut dp = vec![(0, 0)];
let mut ret = 0;
jobs.sort_unstable();
for &(et, st, pf) in &jobs {
let i = match dp.binary_search_by_key(&st, |&(t, _)| t) {
Ok(i) => i,
Err(i) => i - 1,
};
if et != dp.last().unwrap().0 {
dp.push((et, dp[i].1 + pf));
} else if dp.last().unwrap().1 < dp[i].1 + pf {
dp.last_mut().unwrap().1 = dp[i].1 + pf;
}
if ret > dp.last().unwrap().1 {
dp.last_mut().unwrap().1 = ret;
} else {
ret = dp.last().unwrap().1;
}
}
ret
}
}