Minimum Money Required Before Transactions

Minimum Money Required Before Transactions

1. What is this problem about?

This "Hard" difficulty problem involves a sequence of transactions. Each transaction is defined by [cost, cashback]. You must have at least cost money to perform the transaction. After the transaction, your money becomes money - cost + cashback.

The goal is to find the minimum initial money you need to have to guarantee that you can complete all transactions in any possible order. This means you must prepare for the worst-case scenario.

2. Why is this asked in interviews?

Amazon uses this problem to test Greedy logic and Worst-case analysis. It’s tricky because you aren't choosing the order; the order is chosen "against" you. You need to find the "safest" starting amount. Interviewer's look for:

• Impact Analysis: Distinguishing between transactions that lose money (cost > cashback) and those that don't.
• Worst-case Ordering: Realizing that money-losing transactions should generally happen first in a worst-case scenario.
• Mathematical Bound: Finding a single formula to calculate the required amount in $O(n)$.

3. Algorithmic pattern used

The pattern is Greedy Worst-Case Analysis.

1. Transactions that lose money (cost > cashback) contribute to the maximum money "dip." The total money lost from all such transactions is a key component.
2. For each money-losing transaction, the "dip" it causes is its cost, but it also leaves you with cashback. The amount you "lose" forever is cost - cashback.
3. The formula revolves around: Total Loss of all losing transactions + Max(Remaining needed for any single transaction).
4. The "Remaining needed" for a losing transaction i is the cashback[i] (because you already accounted for the cost - cashback in the total loss). For a winning transaction, it's just the cost.

4. Example explanation

Transactions: [[2, 1], [5, 0], [4, 8]]

• [2, 1]: Losing. Loss = 1.
• [5, 0]: Losing. Loss = 5.
• [4, 8]: Winning. No loss. Total Loss = $1 + 5 = 6$.

Worst case: If we do [2, 1] then [5, 0] then [4, 8]:

• To start [2, 1], we need 2. Money becomes 1.
• To start [5, 0], we need 5. We have 1. Need 4 more. Starting = $2+4 = 6$. Money becomes 1.
• To start [4, 8], we need 4. We have 1. Need 3 more. Starting = $6+3 = 9$. If we did them in a different order, we might need less, but we need 10 to cover all orders.

5. Common mistakes candidates make

• Sorting by Cost: Assuming the most expensive transaction always dictates the starting money.
• Ignoring "Winning" Transactions: Thinking only money-losing transactions matter. A winning transaction with a very high cost can still require a high starting amount.
• Simulating All Permutations: $O(n!)$ is impossible for $n=10^5$.

6. Interview preparation tip

In "any order" problems, focus on the "worst order." Usually, this means doing the most "harmful" things (losing money) first and the "least harmful" things (earning money) last.