Problem Description

In this problem, we're given an array transactions that represents a series of transactions. Each transaction is described by two values: a cost and a cashback. The cost is the amount of money that must be paid to perform the transaction, and the cashback is the amount that is returned after the transaction is completed. Our goal is to find the minimum amount of money (money) required before any transaction is made so that it's possible to complete all transactions in any order. The key condition is that at any point, money must be greater than or equal to the cost of the transaction being performed, and after each transaction, money updates to money - cost + cashback.

Intuition

Coming up with a solution involves understanding that there are certain transactions that are riskier in terms of running out of money. Specifically, transactions where cost is greater than cashback are risky because they deplete the pool of money more than any other transaction types. Therefore, one needs to ensure they have enough money to handle the worst-case scenario for these transactions.

Here's the intuition to solve the problem:

1. Calculate the minimum amount of money needed to ensure that you always have enough to cover the cost for the riskier transactions. This is done by finding the sum of the differences between cost and cashback for these transactions (sum(max(0, a - b) for a, b in transactions)).

2. Keep track of the maximum additional money required on top of this sum, which comes from the cashback (or cost if cashback is greater) of the transactions. This step includes the consideration for the transaction with the highest cashback that would minimize the amount of money needed upfront.

The solution iterates through each transaction to calculate these values. The initial sum represents the minimum starting money required to perform all riskier transactions one after another, without the cashback from any transaction dichotomizing the sum. The 'ans' being the maximum helps to keep a running maximum of what that minimum starting money might be considering the cashbacks (or cost, whichever is lower for each of the less risky transactions).

By the end of the iteration, if a transaction's cost is greater than its cashback, the maximum of the sum plus cashback becomes a potential answer. Otherwise, the maximum of the sum plus cost is considered for the less risky transactions. The final ans value is the minimum amount of money required to start with to complete all transactions in any order.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution uses a simple linear scan algorithm, which is efficient since it only needs to iterate through the transactions once. It performs two key calculations:

1. It first calculates the sum of all the positive differences between cost and cashback for each transaction. This is because, for transactions where cost is greater than cashback, you will lose some money, and you need to have enough in reserve to cover these losses. The calculation is done with a generator expression in Python:

   s = sum(max(0, a - b) for a, b in transactions)

   Here, max(0, a - b) ensures that we only sum positive differences, as we do not need extra money upfront for transactions where cashback is greater or equal to cost.

2. The code then determines the maximum additional money required to ensure that transactions can be completed in any order. It iterates through the transactions, comparing each cost and cashback:

   ans = 0
   for a, b in transactions:
       if a > b:
           ans = max(ans, s + b)
       else:
           ans = max(ans, s + a)

   For riskier transactions (a > b), the code considers the possibility of doing this transaction first when the reserve s is at its full. After paying a (cost), you receive b (cashback), so you need at least s + b to start with to do this transaction without going broke.

   For transactions that are not riskier (a <= b), because they refund equal or more than their cost, doing them first would only require enough money to cover the cost, which is a. Thus, for each transaction, you check if s + a (cost) is a new maximum requirement to start with.

The final result in ans is the minimum money required before any transaction that allows you to complete all of them regardless of order. The solution is efficient because it has a time complexity of O(n), where n is the number of transactions, and does not require any additional complex data structures, making use of only iteration and comparison to arrive at the solution.

Example Walkthrough

Let's walk through a small example to better understand the solution approach. Suppose we have the following transactions:

• Transaction 1: cost = 5, cashback = 2
• Transaction 2: cost = 3, cashback = 3
• Transaction 3: cost = 6, cashback = 1

We need to find the minimum amount of money (money) required to complete all these transactions in any order.

First, we calculate the sum of all the positive differences between cost and cashback. In our example:

• For Transaction 1: cost - cashback = 5 - 2 = 3 (positive difference)
• For Transaction 2: cost - cashback = 3 - 3 = 0 (no positive difference)
• For Transaction 3: cost - cashback = 6 - 1 = 5 (positive difference)

We sum these up to get the total sum s:

sum = 3 (from Transaction 1) + 0 (from Transaction 2) + 5 (from Transaction 3) = 8

Next, we iterate over each transaction to decide the maximum additional money required. We start with ans = 0:

• Transaction 1: Since cost > cashback, we consider s + cashback which is 8 + 2 = 10. We update ans to 10 as it's greater than the current ans (0).
• Transaction 2: Since cost <= cashback, we consider s + cost which is 8 + 3 = 11. We update ans to 11 as it's greater than the current ans (10).
• Transaction 3: Since cost > cashback, we consider s + cashback which is 8 + 1 = 9. ans remains 11 as it's greater than 9.

After considering each transaction, the final answer ans is 11. This is the minimum amount of money we need to have upfront before starting any transactions to be able to complete all of them in any order.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code consists of two main parts which contribute to the time complexity:

1. The first part is the sum operation with a generator expression:

   s = sum(max(0, a - b) for a, b in transactions)

   For each transaction in the list transactions, it calculates the maximum between 0 and a - b. This operation has a time complexity of O(N), where N is the number of transactions.

2. The second part is a loop that iterates over all transactions again to determine the maximum money required:

   for a, b in transactions:
       if a > b:
           ans = max(ans, s + b)
       else:
           ans = max(ans, s + a)

   This loop runs for each transaction, making it also O(N).

Since these two parts run sequentially, the overall time complexity is the sum of both parts, which is O(N) + O(N), resulting in a total time complexity of O(N).

Space Complexity

The space complexity of the code is O(1), disregarding the input size. This is because the space used by the variables s and ans is constant and does not depend on the size of the input list transactions. The generator expression also does not create an intermediate list, which keeps the space complexity low.

Learn more about how to find time and space complexity quickly using problem constraints.