Minimum Money Required Before Transactions
<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>
Greedy; Array; Sorting
function minimumMoney(transactions: number[][]): number {
    const s = transactions.reduce((acc, [a, b]) => acc + Math.max(0, a - b), 0);
    let ans = 0;
    for (const [a, b] of transactions) {
        if (a > b) {
            ans = Math.max(ans, s + b);
        } else {
            ans = Math.max(ans, s + a);
        }
    }
    return ans;
}
