Optimal Account Balancing

Hard

Rippling

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Topics:

ArraysGreedy AlgorithmsRecursion

You are given an array of transactions transactions where transactions[i] = [from_i, to_i, amount_i] indicates that the person with ID from_i gave amount_i $ to the person with ID to_i.

Return the minimum number of transactions required to settle the debt between a set of people.

Note:

If the total debt is not 0, it means that some transactions have been overlooked. In that case, output -1.

Example 1:

Input: transactions = [[0,1,10],[2,0,5]]
Output: 2
Explanation:
PersonId 0 gave personId 1, 10$.
PersonId 2 gave personId 0, 5$.

Two transactions are needed. One person needs to give another one 5$ and
a second person needs to give the first one 5$.

Example 2:

Input: transactions = [[0,1,10],[1,0,1],[1,2,5],[2,0,5]]
Output: 1
Explanation:
PersonId 0 gave personId 1, 10$.
PersonId 1 gave personId 0, 1$.
PersonId 1 gave personId 2, 5$.
PersonId 2 gave personId 0, 5$.

Therefore,
person 1 only need to give person 0 9$.
So only one transaction is needed.

Constraints:

1 <= transactions.length <= 8
transactions[i].length == 3
0 <= from_i, to_i < 20
from_i != to_i
1 <= amount_i <= 100

Optimal Account Balancing

Hard

Rippling

3 views

Topics:

ArraysGreedy AlgorithmsRecursion

You are given an array of transactions transactions where transactions[i] = [from_i, to_i, amount_i] indicates that the person with ID from_i gave amount_i $ to the person with ID to_i.

Return the minimum number of transactions required to settle the debt between a set of people.

Note:

If the total debt is not 0, it means that some transactions have been overlooked. In that case, output -1.

Example 1:

Input: transactions = [[0,1,10],[2,0,5]]
Output: 2
Explanation:
PersonId 0 gave personId 1, 10$.
PersonId 2 gave personId 0, 5$.

Two transactions are needed. One person needs to give another one 5$ and
a second person needs to give the first one 5$.

Example 2:

Input: transactions = [[0,1,10],[1,0,1],[1,2,5],[2,0,5]]
Output: 1
Explanation:
PersonId 0 gave personId 1, 10$.
PersonId 1 gave personId 0, 1$.
PersonId 1 gave personId 2, 5$.
PersonId 2 gave personId 0, 5$.

Therefore,
person 1 only need to give person 0 9$.
So only one transaction is needed.

Constraints:

1 <= transactions.length <= 8
transactions[i].length == 3
0 <= from_i, to_i < 20
from_i != to_i
1 <= amount_i <= 100

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:

What are the possible ranges for the 'from', 'to', and 'amount' values in the transactions?
Can the 'amount' in a transaction ever be zero or negative?
Is it possible for the same (from, to) pair to appear multiple times in the transactions list? If so, how should those amounts be handled?
Is the list of transactions guaranteed to be such that a balanced state is always achievable? What should I return if it is not?
Are the 'from' and 'to' values indices into a separate list of people or are they unique identifiers that can be used directly, for example integers or strings?

Brute Force Solution

Approach

The goal is to make sure everyone's debts are settled using the fewest number of transactions possible. The brute force approach to solving this problem involves trying every single possible combination of transactions to see which one uses the least number of transactions overall.

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

First, figure out how much money each person owes or is owed. Some people will be in debt, and others will have extra money.
Then, start trying different ways to match up people who owe money with people who are owed money. Think of it like trying all possible exchanges between pairs or groups of people.
For each possible matching, check if the debts are actually getting settled. Is the total amount owed being balanced out by the total amount being received?
Keep track of the number of transactions needed for each possible way of settling the debts.
Compare all the different ways you found to settle the debts, and pick the one that uses the fewest transactions.

Code Implementation

def min_transactions_brute_force(transactions):

    net_balances = {}
    for transaction in transactions:
        debtor = transaction[0]
        creditor = transaction[1]
        amount = transaction[2]
        net_balances[debtor] = net_balances.get(debtor, 0) - amount
        net_balances[creditor] = net_balances.get(creditor, 0) + amount

    balances = list(net_balances.values())
    
    def settle_debts(balances, transaction_count):
        if not balances:
            return transaction_count

        for i in range(len(balances)):
            if balances[i] == 0:
                continue

            temp_balances = balances[:]

            # Iterate through the remaining non-zero balances
            for j in range(i + 1, len(balances)):        
                if balances[j] == 0:
                    continue

                temp_balances = balances[:]
                temp_balances[i] -= balances[j]
                temp_balances[j] = 0

                # This is a core decision point. 
                # Recursively try all the settlement combinations.
                min_count = settle_debts(temp_balances, transaction_count + 1)
                return min_count

        return float('inf')
    
    # Start the recursive process
    # by initiating transaction_count to zero
    minimum_transaction_count = settle_debts(balances, 0)

    if minimum_transaction_count == float('inf'):
        return 0 # All balances must have been zero

    return minimum_transaction_count

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n!) – The problem explores all possible transactions between debtors and creditors. In the worst-case scenario, we might need to consider all permutations of how n people (or groups of people with net debt/credit) can transact with each other to settle debts. Generating all these permutations dominates the runtime. The number of such permutations grows factorially with n, resulting in O(n!) complexity. This is because for n people, there are n! (n factorial) ways to arrange them, and the brute force approach considers these arrangements when matching debtors to creditors.

Space Complexity

O(N) – The algorithm identifies individuals who owe or are owed money, and the plain English explanation implies storing these debts in some manner. This suggests a data structure, likely a list or hash map, to store the net balance for each person. In the worst-case scenario, each person has a unique balance, meaning we would store balances for N people, where N is the number of people involved in the transactions. Therefore, the auxiliary space required is proportional to the number of people, resulting in O(N) space complexity.

Optimal Solution

Approach

The most efficient way to solve this problem is to simplify it by focusing on what each person ultimately owes or is owed. We can then iteratively settle these debts by pairing up creditors and debtors to cancel out as much debt as possible in each transaction. By strategically pairing people, we minimize the total number of transactions needed.

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

First, calculate the net amount each person has after all the transactions. Some people will have a positive balance (they are owed money), and others will have a negative balance (they owe money). Some will have a zero balance (they are settled).
Combine all the positive and negative balance values, ignoring the zeros. The goal is to eliminate these balances.
Pick a person who is owed money and a person who owes money. Ideally, pick the largest debt and credit for maximum efficiency.
Figure out the maximum amount that can be settled between these two people. It will be the smaller of the amount owed and the amount to be received.
Record this transaction. The person who owed money pays the person who was owed, up to the limit calculated in the previous step.
Adjust the balances of both people to reflect the transaction. One person's debt or credit might be completely settled, or both accounts might be reduced.
If there are any remaining debts or credits, repeat the process from step 3. Otherwise, you're done!
The total number of transactions recorded is the minimum number of transactions needed to settle all debts.

Code Implementation

def min_transactions(transactions):
    person_balances = {}
    for transaction in transactions:
        payer = transaction[0]
        receiver = transaction[1]
        amount = transaction[2]
        person_balances[payer] = person_balances.get(payer, 0) - amount
        person_balances[receiver] = person_balances.get(receiver, 0) + amount

    balances = list(person_balances.values())
    balances = [balance for balance in balances if balance != 0]

    def settle_debts(balances):
        number_of_transactions = 0
        while balances:
            # Find the max credit and max debit to settle
            max_credit = max(balances)
            max_debit = min(balances)

            amount_to_settle = min(abs(max_debit), max_credit)

            # Settle as much as possible
            balances.remove(max_credit)
            balances.remove(max_debit)

            if max_credit - amount_to_settle != 0:
                balances.append(max_credit - amount_to_settle)

            if max_debit + amount_to_settle != 0:
                balances.append(max_debit + amount_to_settle)

            number_of_transactions += 1

        return number_of_transactions

    # Balances represents all net debts and credits.
    return settle_debts(balances)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^2) – The described algorithm involves iterating through the balances to find creditors and debtors. In the worst-case scenario, for each person (up to n), we might have to compare them with every other person to find a suitable match for debt settlement. This pairwise comparison results in a nested loop structure where, for each of the n individuals, we potentially examine the remaining (n-1) individuals. Therefore, the total number of operations grows proportionally to n * (n-1), which simplifies to O(n^2).

Space Complexity

O(N) – The dominant space usage comes from storing the net amounts each person has after all transactions. In the worst case, each person could have a unique balance, leading to an auxiliary data structure (likely a list or hash map implicitly mentioned when the explanation states we 'combine all the positive and negative balance values') to store these balances, where N is the number of people involved in the transactions. The algorithm also uses variables to store the largest debt and credit, but these use constant space. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty transaction list	Return 0 if the transaction list is null or empty, as there are no debts to settle.
Transaction list with a single transaction	Return 1 as a single transaction is already the minimum required to settle that single debt.
All transactions result in a net balance of zero for all individuals.	Return 0 since all accounts are already balanced and no transactions are needed.
Large transaction amounts leading to potential integer overflow when calculating balances.	Use a data type that can accommodate large numbers, such as long, to prevent integer overflow.
Transactions forming a cycle (e.g., A owes B, B owes C, C owes A)	The balance calculation and zeroing-out approach correctly handles cycles by iteratively reducing debts.
A single person owes multiple people, or multiple people owe a single person.	The balance calculation correctly aggregates debts to and from each person.
Transactions with identical (from, to) pairs but different amounts (representing multiple debts between the same two people).	The balance calculation correctly aggregates the amounts for duplicate (from, to) pairs, treating them as a single consolidated debt.
Large number of people and transactions leading to memory constraints.	The memory usage should scale linearly with the number of people, which should generally be manageable for reasonable problem sizes.