Problem Description

In this problem, we are provided with a list of financial transactions among a group of people, where each person has a unique ID. Each transaction is represented as a list with three integers [from_i, to_i, amount_i] signifying that the person with ID from_i has given the amount amount_i to the person with ID to_i. The objective is to determine the minimum number of transactions needed to settle all debts in the system. In other words, we want to find the smallest set of transactions so that everybody ends up with a balance of zero without considering the original transactions.

Flowchart Walkthrough

Let's analyze leetcode 465. Optimal Account Balancing using the Flowchart. Here is a step-by-step analysis of the problem:

1. Is it a graph?

   • Yes: Though it may not look immediately like a traditional graph problem, the essence of balancing accounts can be seen as creating and balancing transactions between nodes (accounts).

2. Does the problem have small constraints?

   • Yes: While not strictly limited in number, the real-world scenarios typically involve a manageable number of accounts, suggesting that we might encounter small constraints for practical purposes.

3. Brute force / Backtracking?

   • Yes: Given the complexity of various transactions and the requirement to minimize the transaction count, brute force or backtracking can be highly suitable. This is because every transaction creates a new state of balances, and we may need to backtrack to find the optimal series of transactions.

Conclusion: The flowchart leads us to consider using a backtracking approach to explore different ways transactions can balance the accounts efficiently while minimizing the total number of transactions needed. Such a method systematically tries different transactions, possibly undoing them (backtrack) if they do not lead toward the solution.

Intuition

The intuition behind the solution is to first simplify the problem by calculating the net balance of each person after all the transactions. A net balance is the amount of money a person owes (negative balance) or is owed (positive balance). If a person has a net balance of zero, no further action is needed for them, so we ignore them in the settlement process.

To settle the debts, we aim to find the fewest number of transactions that would balance these net amounts. This becomes a combinatorial optimization problem where we explore different ways to match debtors (people with a negative balance) with creditors (people with a positive balance) in a minimal fashion. Essentially, it's a variation of the subset sum problem, which is computationally challenging (NP-Complete) due to the number of combinations to consider.

We use dynamic programming to manage the complexity. We represent each subset of people using a bitmask, where the i-th bit represents whether the i-th person is included in the subset. Our dynamic programming array f keeps track of the minimum number of transactions needed for each subset of people. The value f[i] contains the optimal (minimum) number of transactions needed to settle the debts among the subset of people represented by the bitmask i.

By iterating through all possible subsets, we try to construct solutions from combinations of smaller subsets that sum up to zero, meaning their debts can be settled among themselves. We use bit manipulation to iterate through subsets and to compute the number of bits set (number of people involved) in a state. The bit count of a state minus one gives us the correct number of transactions, assuming a direct transaction could settle the debts among people in that state.

The final answer is then found in f[-1], which represents the minimum number of transactions needed to settle all debts among all people, encoded in the bitmask with all bits set.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The solution relies on dynamic programming, bit manipulation, and understanding of financial transactions to reduce an optimization problem into a solvable algorithm. Let's walk through it step-by-step with reference to the given Python code:

1. Preparing the Net Balance: We initialize a defaultdict(int) named g to store the net balance of every individual. As we iterate through each transaction [from_i, to_i, amount_i] in transactions, we update the net balance by subtracting amount_i from from_i's balance and adding it to to_i's balance. This will ensure that the net balance reflects how much each person is in debt (negative balance) or is owed (positive balance).

2. Filtering Zero Balances: We filter out individuals with a zero net balance since they do not need to be involved in any further transactions. This is done by creating a list nums containing only non-zero balances.

3. Initializing the Dynamic Programming Array: We define a one-dimensional DP array f with length 1 << m, where m is the number of non-zero net balances (the length of nums). The notation 1 << m means we are using a bitmask to represent all possible subsets of these m balances. We initialize all values in f to infinity (inf), except f[0] which is set to 0 since no transactions are needed if no one is involved.

4. Finding the Minimum Transactions: We iterate over all subsets of nums by using the variable i to represent a bitmask. For each subset i, we calculate the total sum s of the balances in that subset. If the sum is zero (s == 0), it means this subset can balance itself out with no debt remaining.

5. Setting Base Case for Valid Subsets: If the i-th subset's sum is 0, we calculate the number of required transactions as one less than the number of set bits in i. This is because, if n people are in a subset, n - 1 transactions are sufficient to settle their debts (everyone settles with one person).

6. Optimizing the Transactions: For each valid subset i, we look for pairs of disjoint subsets j and i ^ j (where ^ is the XOR operation) that combine to form i and have a sum of zero. We use bit manipulation (j - 1) & i to iterate through all those potential pairs. The goal is to find the pair that minimizes the sum of f[j] + f[i ^ j], which represents the minimum transactions needed to settle the subsets j and i ^ j separately.

7. Final Answer: The last element in f, f[-1], holds the smallest number of transactions needed to settle all debts, which is what the function minTransfers returns.

In summary, this approach uses a combination of net balance calculation, filtering, dynamic programming, and bit manipulation to efficiently find the minimum number of transactions needed to settle debts across a group of people.

Example Walkthrough

Let's use a small example to illustrate the solution approach described above. Suppose we have the following set of transactions between three people with IDs 1, 2, and 3:

Transactions: [[1, 2, 10], [2, 3, 5], [3, 1, 5]]

This means that person 1 gave 10 to person 2, person 2 gave 5 to person 3, and person 3 gave 5 to person 1.

Step 1: Preparing the Net Balance

We calculate the net balance of each individual:

• Person 1: (received 5) - (gave 10) = -5
• Person 2: (received 10) - (gave 5) = 5
• Person 3: (received 5) - (gave 5) = 0

Step 2: Filtering Zero Balances

Person 3 has a net balance of 0, so they are excluded from further transactions:

nums: [-5, 5]

Step 3: Initializing the Dynamic Programming Array

There are 2 people with non-zero net balances, so our DP array f will have 1 << 2 (which is 4) elements:

f: [0, inf, inf, inf]

Step 4: Finding the Minimum Transactions

We iterate over all subsets of nums:

• Subset represented by bitmask 00 has sum 0. We already set f[0] to 0 because no people are involved.
• Subset 01 has only person 1, who cannot settle the debt alone.
• Subset 10 has only person 2, who also cannot settle the debt alone.
• Subset 11 has both person 1 and 2. The sum is 0, so they can settle among themselves.

Step 5: Setting Base Case for Valid Subsets

Since the sum for the subset 11 is 0, we can settle the debt with one transaction:

f[11] = Number of people in subset - 1 = 2 - 1 = 1

So, f[3] (binary 11 is 3 in decimal), is updated to 1.

Step 6: Optimizing the Transactions

This step involves disjoint pairs but, in our case, since we have just 2 non-zero net balances, there is only one valid subset (11) that balances to zero. So we skip the optimization as we've already found the solution.

Step 7: Final Answer

The final answer, found in f[-1] (or f[3] since we only have 4 entries and Python allows negative indexing), is the minimum number of transactions needed which is 1. One person pays the other, and all debts are settled.

By this example, we can see how the described dynamic programming solution would work on a small scale and calculate the minimum number of transactions to settle debts.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code's time complexity is primarily determined by the nested for loops.

1. Iteration over transactions: O(N), where N is the number of transactions.

2. Iteration over powerset of debts: This loop iterates over all subsets of a set of m debts, leading to 2^m iterations, where m is the number of unique debts.

3. Inner loop to calculate sum of debts subsets: O(m) per subset, since it iterates over all m debts to identify sums that cancel out to zero.

4. Innermost while loop for calculating f[i]: The loop will run in the worst-case scenario 2^m times, for each of the combinations from the second iteration.

Consequently, the time complexity is O(N) + O(m * 2^m) + O(m * (2^m)) which simplifies to O(N + m * 2^m), where m is the number of non-zero individual net balances. Since m can be at most equivalent to N, in the worst case, the m * 2^m term dominates, making the time complexity O(m * 2^m).

Space Complexity

The space complexity is influenced by:

1. The hashmap (g) to store the balances: O(M), where M is the number of unique people involved.

2. The list nums of non-zero balances: O(m).

3. The list f of size 2^m to keep track of minimum transfers for each subset of debts: O(2^m).

Hence, the space complexity is O(M + m + 2^m). However, since the 2^m term is the most significant, it simplifies to O(2^m).

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