## 826. Most Profit Assigning Work

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]`

## Rust Solution

``````struct Solution;
use std::collections::BTreeMap;

impl Solution {
fn max_profit_assignment(difficulty: Vec<i32>, profit: Vec<i32>, worker: Vec<i32>) -> i32 {
let n = difficulty.len();
let mut btm: BTreeMap<i32, i32> = BTreeMap::new();
for i in 0..n {
let v = btm.entry(difficulty[i]).or_default();
*v = profit[i].max(*v);
}
let mut prev = 0;
for (_, v) in btm.iter_mut() {
if prev > *v {
*v = prev;
}
prev = *v;
}
let mut res = 0;
for w in worker {
res += *btm.range(0..=w).rev().map(|(_, v)| v).next().unwrap_or(&0);
}
res
}
}

#[test]
fn test() {
let difficulty = vec![2, 4, 6, 8, 10];
let profit = vec![10, 20, 30, 40, 50];
let worker = vec![4, 5, 6, 7];
let res = 100;
assert_eq!(
Solution::max_profit_assignment(difficulty, profit, worker),
res
);
}
``````

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