# Solving Leetcode Interviews in Seconds with AI: Minimum Money Required Before Transactions


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2412" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a 0-indexed 2D integer array transactions, where transactions[i] = [costi, cashbacki]. The array describes transactions, where each transaction must be completed exactly once in some order. At any given moment, you have a certain amount of money. In order to complete transaction i, money >= costi must hold true. After performing a transaction, money becomes money - costi + cashbacki. Return the minimum amount of money required before any transaction so that all of the transactions can be completed regardless of the order of the transactions.   Example 1:  Input: transactions = [[2,1],[5,0],[4,2]] Output: 10 Explanation: Starting with money = 10, the transactions can be performed in any order. It can be shown that starting with money < 10 will fail to complete all transactions in some order.  Example 2:  Input: transactions = [[3,0],[0,3]] Output: 3 Explanation: - 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.    Constraints:  1 <= transactions.length <= 105 transactions[i].length == 2 0 <= costi, cashbacki <= 109  

	# Explanation
	Here's the solution:

*   **Separate and Sort:** The core idea is to separate transactions into two groups: those where `cost > cashback` and those where `cost <= cashback`. Transactions where `cost > cashback` must be performed first, and should be sorted in descending order of the difference `cost - cashback`. Transactions where `cost <= cashback` can be performed later and can be sorted in ascending order of cost
*   **Greedy Execution:** Simulate performing the transactions in the determined order (first the sorted `cost > cashback` transactions, then the sorted `cost <= cashback` transactions) and track the minimum initial money required.
*   **Handle `cost <= cashback` transactions:** Consider `cost <= cashback` transactions after `cost > cashback` transactions, sorting them in ascending order of their costs. This minimizes the immediate money needed and ensures all such transactions can be performed subsequently

*   **Time Complexity:** O(N log N), where N is the number of transactions (due to sorting).
*   **Space Complexity:** O(N) in the worst-case scenario, where N is the number of transactions (due to the `loss` and `profit` lists).

	
	# Code
	```python
	def minimumMoney(transactions):
    """
    Calculates the minimum amount of money required to complete all transactions.

    Args:
        transactions: A 2D integer array representing transactions, where transactions[i] = [costi, cashbacki].

    Returns:
        The minimum amount of money required.
    """

    loss = []
    profit = []
    for cost, cashback in transactions:
        if cost > cashback:
            loss.append((cost, cashback))
        else:
            profit.append((cost, cashback))

    loss.sort(key=lambda x: x[0] - x[1], reverse=True)
    profit.sort(key=lambda x: x[0])

    money = 0
    current_money = 0

    for cost, cashback in loss:
        if current_money < cost:
            money += cost - current_money
            current_money = cost
        current_money = current_money - cost + cashback

    for cost, cashback in profit:
        if current_money < cost:
            money += cost - current_money
            current_money = cost
        current_money = current_money - cost + cashback

    return money
	```