<p>You are given a <strong>0-indexed</strong> 2D integer array <code><font face="monospace">transactions</font></code>, where <code>transactions[i] = [cost<sub>i</sub>, cashback<sub>i</sub>]</code>.</p>

<p>The array describes transactions, where each transaction must be completed exactly once in <strong>some order</strong>. At any given moment, you have a certain amount of <code>money</code>. In order to complete transaction <code>i</code>, <code>money &gt;= cost<sub>i</sub></code> must hold true. After performing a transaction, <code>money</code> becomes <code>money - cost<sub>i</sub> + cashback<sub>i</sub></code>.</p>

<p>Return<em> the minimum amount of </em><code>money</code><em> required before any transaction so that all of the transactions can be completed <strong>regardless of the order</strong> of the transactions.</em></p>

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

<pre>
<strong>Input:</strong> transactions = [[2,1],[5,0],[4,2]]
<strong>Output:</strong> 10
<strong>Explanation:
</strong>Starting with money = 10, the transactions can be performed in any order.
It can be shown that starting with money &lt; 10 will fail to complete all transactions in some order.
</pre>

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

<pre>
<strong>Input:</strong> transactions = [[3,0],[0,3]]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
- If transactions are in the order [[3,0],[0,3]], the minimum money required to complete the transactions is 3.
- If transactions are in the order [[0,3],[3,0]], the minimum money required to complete the transactions is 0.
Thus, starting with money = 3, the transactions can be performed in any order.
</pre>

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

<ul>
	<li><code>1 &lt;= transactions.length &lt;= 10<sup>5</sup></code></li>
	<li><code>transactions[i].length == 2</code></li>
	<li><code>0 &lt;= cost<sub>i</sub>, cashback<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minimumMoney(self, A: List[List[int]]) -> int:
        
        # Greedy Approach --> 
        # Splist transcations into two groups and sort according to the order.

        L, E = [a for a in A if a[0] > a[1]], [a for a in A if a[0] <= a[1]]

        L.sort(key = lambda x : x[1])

        # budget: minimum budget we have met.
        # curr: current budget we have.
        budget, curr = 0, 0

        for a, b in L:
            curr -= a
            budget = min(budget, curr)
            curr += b

        if E:
            curr -= max(e[0] for e in E)
            budget =  min(budget, curr)

        return -budget


        # LEE's Solution --> Sorting Not needed
        res = v = 0
        for i,j in A:
            res += max(0, i - j)
            v = max(v, min(i, j))
        return res + v
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