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

<p>You&#39;re given the <code>startTime</code>, <code>endTime</code> and <code>profit</code> arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.</p>

<p>If you choose a job that ends at time <code>X</code> you will be able to start another job that starts at time <code>X</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png" style="width: 380px; height: 154px;" /></strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png" style="width: 600px; height: 112px;" /> </strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png" style="width: 400px; height: 112px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
<strong>Output:</strong> 6
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= startTime.length == endTime.length == profit.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= startTime[i] &lt; endTime[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= profit[i] &lt;= 10<sup>4</sup></code></li>
</ul>

def go(*, times_by_start, times_by_end, s_keys, e_keys, idx, level, dp) -> tuple[int, int]:

    if (hit := dp.get(key := (
        idx,
    ))):
        return hit
    
    def find_first_succ_idx(end, l, r):
        if l == r:
            return l
        mid_idx = l + (r - l) // 2
        mid_val = s_keys[mid_idx]
        if end <= mid_val:
            return find_first_succ_idx(end, l, mid_idx)
        return find_first_succ_idx(end, mid_idx + 1, r)
        
    n = len(s_keys)
    if idx >= n:
        dp[key] = 0x3f3f3f3f, 0
        return dp[key]
    tail_end, tail_result = go(
        times_by_start=times_by_start,
        times_by_end=times_by_end,
        s_keys=s_keys,
        e_keys=e_keys,
        idx=idx + 1,
        level=level + 1,
        dp=dp,
    )
    s_key = s_keys[idx]
    res_end, res_result = tail_end, tail_result
    for start, end, profit in times_by_start[s_key]:
        if end <= tail_end:
            taken_result = tail_result + profit
            if taken_result > res_result:
                res_result = taken_result
                res_end = start
        else:
            if (t_idx := find_first_succ_idx(end, idx + 1, len(s_keys))) is not None:
                succ_end, succ_result = go(
                    times_by_start=times_by_start,
                    times_by_end=times_by_end,
                    s_keys=s_keys,
                    e_keys=e_keys,
                    idx=t_idx,
                    level=level + 1,
                    dp=dp,
                )
                taken_result = succ_result + profit
                if taken_result > res_result:
                    res_result = taken_result
                    res_end = start
    dp[key] = res_end, res_result
    return dp[key]

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        times = tuple(zip(startTime, endTime, profit))
        times_by_start = defaultdict(list)
        times_by_end = defaultdict(list)
        for start, end, profit in times:
            times_by_start[start].append((start, end, profit))
            times_by_end[end].append((start, end, profit))
        res = go(
            times_by_start=times_by_start,
            times_by_end=times_by_end,
            s_keys=sorted(times_by_start.keys()),
            e_keys=sorted(times_by_end.keys()),
            idx=0,
            level=0,
            dp={},
        )
        return res[1]
        
91615