Minimum Money Required Before Transactions

Hard

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Greedy Algorithms

You are given a 0-indexed 2D integer array transactions, where transactions[i] = [cost_i, cashback_i].

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 >= cost_i must hold true. After performing a transaction, money becomes money - cost_i + cashback_i.

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 <= cost_i, cashback_i <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the transaction amounts be negative, zero, or only positive, and what is the range of possible values for each transaction?
If it's impossible to complete all transactions, what should I return?
Is the input array guaranteed to be non-empty, and if not, how should I handle an empty input?
Does the order of transactions in the input array matter, or can I reorder them for the calculation?
Could you provide a more concrete example to illustrate the expected behavior with a mix of positive and negative transaction amounts?

Brute Force Solution

Approach

We need to find the smallest amount of money needed at the start so you never go broke while paying for several transactions. The brute-force method tries out every possible starting amount, one by one, until we find one that works.

Here's how the algorithm would work step-by-step:

Start with a very small amount of money as your initial guess.
Go through each transaction, subtracting the cost from your current money.
If, at any point, your money goes below zero, that starting amount was not enough.
Increase your initial guess and start over with the transactions again.
Keep doing this, increasing the initial guess each time, until you find a starting amount where you can pay for all transactions without ever running out of money.
That starting amount is the minimum money required.

Code Implementation

def minimum_money_brute_force(transactions):
    initial_money = 0

    while True:
        sufficient_funds = True
        current_money = initial_money

        # Iterate through the transactions
        for transaction_cost in transactions:

            current_money -= transaction_cost

            # Check if we went broke
            if current_money < 0:
                sufficient_funds = False
                break

        # We found a starting amount that works
        if sufficient_funds:
            return initial_money

        # Increase the initial guess
        initial_money += 1

Big(O) Analysis

Time Complexity

O(M*n) – The described brute-force solution involves an outer loop that iteratively increases the initial money, let's call the maximum initial money needed M. The inner loop iterates through the n transactions to check if a given initial money is sufficient. The worst-case time complexity is therefore proportional to M (the number of attempts to find the minimum money) multiplied by n (the number of transactions to process in each attempt). Thus, the overall time complexity is O(M*n).

Space Complexity

O(1) – The brute-force method described only uses a single variable to represent the current money and iterates through the transactions. The algorithm does not create any auxiliary data structures like arrays, lists, or hash maps to store intermediate results. Therefore, the amount of extra memory used remains constant regardless of the number of transactions (N). The auxiliary space complexity is O(1).

Optimal Solution

Approach

The key idea is to work backward from the end to figure out the minimum starting amount. We only care about the scenarios where we end up in debt, so we focus on covering those.

Here's how the algorithm would work step-by-step:

Start by assuming you need zero money initially.
Go through the transactions in reverse order, from the last transaction to the first.
For each transaction, check if it will put you in debt (result in a negative balance).
If a transaction puts you in debt, increase your initial money by the amount needed to avoid the debt.
Continue this process until you've checked all transactions in reverse order.
The final amount of initial money is the minimum amount required.

Code Implementation

def minimum_money_required(transactions):
    minimum_starting_money = 0
    current_balance = 0

    # Iterate transactions in reverse order
    for transaction_amount in reversed(transactions):

        current_balance += transaction_amount

        # Check if balance goes below zero.
        if current_balance < 0:

            # Need more initial money to cover the debt
            minimum_starting_money += abs(current_balance)
            current_balance = 0

    return minimum_starting_money

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the transactions array once, but in reverse order. The input size, n, represents the number of transactions. Inside the loop, there are only constant-time operations like checking if the current balance becomes negative and potentially updating the initial money. Therefore, the time complexity is directly proportional to the number of transactions, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm iterates through the transactions in reverse order but doesn't store any additional data structures that scale with the input size (N, where N is the number of transactions). It only uses a few variables to keep track of the current money and the initial money needed. Therefore, the auxiliary space used is constant, independent of the number of transactions.

Edge Cases

Case	How to Handle
Empty transactions array	Return 0 as no money is needed if there are no transactions.
All transactions are positive (gains)	Return 0 as you always have enough money to start.
All transactions are negative (losses)	Sum all negative values and take the absolute value as the starting amount needed.
Integer overflow with large transaction values	Use a data type (e.g., long) that can accommodate large sums to prevent overflow.
Transactions resulting in a zero balance at some point	The algorithm should correctly track the minimum balance reached, even if it's zero.
Array containing very large positive and negative numbers that offset each other to 0	The iterative sum will still correctly identify the maximum negative balance requiring upfront money.
The absolute sum of losses is greater than the maximum possible integer value.	Throw an exception, return an error code, or switch to BigInteger if supported by the language.
Array contains only zero values.	Return 0 as no starting money is needed with zero-value transactions.

Empty transactions array

How to Handle:

Return 0 as no money is needed if there are no transactions.

All transactions are positive (gains)

How to Handle:

Return 0 as you always have enough money to start.

All transactions are negative (losses)

How to Handle:

Sum all negative values and take the absolute value as the starting amount needed.

Integer overflow with large transaction values

How to Handle:

Use a data type (e.g., long) that can accommodate large sums to prevent overflow.

Transactions resulting in a zero balance at some point

How to Handle:

The algorithm should correctly track the minimum balance reached, even if it's zero.

Array containing very large positive and negative numbers that offset each other to 0

How to Handle:

The iterative sum will still correctly identify the maximum negative balance requiring upfront money.

The absolute sum of losses is greater than the maximum possible integer value.

How to Handle:

Throw an exception, return an error code, or switch to BigInteger if supported by the language.

Array contains only zero values.

How to Handle:

Return 0 as no starting money is needed with zero-value transactions.