<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>

from itertools import chain


class Solution:
    def jobScheduling(self, startTime: list[int], endTime: list[int], profit: list[int]) -> int:
        # Step 1: Combine startTime, endTime, and profit into a single list and sort by endTime
        jobs = sorted(zip(startTime, endTime, profit), key=lambda x: x[1])

        # Step 2: Extract all unique times (for coordinate compression)
        times = sorted(frozenset(chain(startTime, endTime)))

        # Step 3: Coordinate compression, mapping original times to a smaller range
        time_map = {time: i for i, time in enumerate(times, 1)}  # Using 1-based index

        # Step 4: Initialize BIT
        bit = BinaryIndexedTree(len(times), max)

        # Step 5: Process jobs and use BIT to find max profit
        max_profit = 0
        for s, e, p in jobs:
            # Find the maximum profit that can be obtained before the job's start time
            start_idx = time_map[s]  # Compressed index for start time
            end_idx = time_map[e]  # Compressed index for end time

            # Max profit before current job's start time
            current_profit = bit.query(start_idx)

            # Calculate total profit including the current job
            total_profit = current_profit + p

            # Update BIT with the new profit at the job's end time
            bit.update(end_idx, total_profit)

            # Keep track of the maximum profit globally
            max_profit = max(max_profit, total_profit)

        return max_profit


class BinaryIndexedTree:
    def __init__(self, size, function):
        self.size = size
        self.function = function
        self.tree = [0] * (size + 1)

    def __len__(self):
        return self.size

    def query(self, index):
        max_val = 0
        while index > 0:
            max_val = self.function(self.tree[index], max_val)
            index -= index & -index
        return max_val

    def update(self, index, value):
        while index <= self.size:
            self.tree[index] = self.function(value, self.tree[index])
            index += index & -index
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