LeetCode 465: Optimal Account Balancing Solution in Python – A Step-by-Step Guide

What Is LeetCode 465: Optimal Account Balancing?

In LeetCode 465: Optimal Account Balancing, you’re given a list of transactions as tuples [from_id, to_id, amount], representing debts between people (e.g., [0,1,10] means person 0 owes person 1 $10). Your task is to find the minimum number of transactions to settle all debts, where each transaction moves money directly between two people to reduce net balances to zero. For example, with [[0,1,10], [1,0,5], [1,2,5], [2,0,5]], the minimum is 1 (person 1 pays person 0 $5). It’s like streamlining a tangled web of IOUs into the fewest handshakes.

Problem Statement

• Input: transactions (List[List[int]])—list of [from_id, to_id, amount].
• Output: int—minimum number of transactions to balance accounts.
• Rules:
• Each transaction reduces net debt.
• All accounts must balance (net = 0).
• IDs are 0-based integers.

Constraints

• 1 <= transactions.length <= 8.
• transactions[i].length == 3.
• 0 <= from_id, to_id < 12.
• 1 <= amount <= 100.

Examples to Get Us Started

• Input: transactions = [[0,1,10], [1,0,5], [1,2,5], [2,0,5]]
• Output: 1 (Person 1 pays person 0 $5).
• Input: transactions = [[0,1,10], [2,0,5]]
• Output: 2 (Two separate debts).
• Input: transactions = [[0,1,2]]
• Output: 1 (One transaction).

Understanding the Problem: Settling the Debts

To solve LeetCode 465: Optimal Account Balancing in Python, we need to minimize transactions by reducing net debts to zero. A naive approach—trying all possible transaction sequences—is O(n!) or worse, infeasible even with n ≤ 8. The key? Compute net balances and use DFS to pair debts optimally, pruning redundant paths. We’ll explore:

• Best Solution (DFS with Pruning): O(n * 2^n) time, O(n) space—fast and smart.
• Alternative Solution (Brute Force): O(n!) time, O(n) space—simple but slow.

Let’s dive into the DFS solution—it’s the treasurer’s ledger we need.

Best Solution: DFS with Pruning

Why This Is the Best Solution

The DFS with pruning approach is the top pick because it’s O(n * 2^n) time and O(n) space, efficiently finding the minimum transactions by exploring debt pairings with backtracking and skipping zeros. It computes net balances, then uses DFS to match positive and negative debts, minimizing moves. It’s like sorting through IOUs and finding the shortest handshake chain—clever and optimized!

How It Works

Here’s the strategy:

• Step 1: Compute net balances for each person:
• For [from_id, to_id, amount], subtract from from_id, add to to_id.
• Step 2: Filter non-zero balances into a list (debts).
• Step 3: DFS on debts:
• Skip zeros, pick a debt, pair with opposite sign debts.
• Recurse with updated list, track min transactions.
• Step 4: Return minimum moves.
• Why It Works:
• Non-zero debts must balance to zero.
• DFS explores all valid pairings, pruning optimizes.

Step-by-Step Example

Example: transactions = [[0,1,10], [1,0,5], [1,2,5], [2,0,5]]

• Net Balances:
• 0: -10 + 5 - 5 = -10.
• 1: +10 - 5 - 5 = 0.
• 2: +5 + 5 = 10.
• Debts: [-10, 10].
• DFS:
• Start: [-10, 10], moves = 0.
• Pick -10, pair with 10: [0], moves = 1.
• Base case: all 0s, moves = 1.
• Result: 1.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def minTransfers(self, transactions: List[List[int]]) -> int:
        # Step 1: Compute net balances
        balance = {}
        for from_id, to_id, amount in transactions:
            balance[from_id] = balance.get(from_id, 0) - amount
            balance[to_id] = balance.get(to_id, 0) + amount

        # Step 2: Filter non-zero debts
        debts = [val for val in balance.values() if val != 0]
        if not debts:
            return 0

        # Step 3: DFS with pruning
        def dfs(start, debts, moves):
            # Skip zeros
            while start < len(debts) and debts[start] == 0:
                start += 1
            if start >= len(debts):
                return moves

            min_moves = float('inf')
            curr_debt = debts[start]
            for i in range(start + 1, len(debts)):
                if debts[i] * curr_debt < 0:  # Opposite signs
                    # Try pairing
                    debts[i] += curr_debt
                    min_moves = min(min_moves, dfs(start + 1, debts, moves + 1))
                    debts[i] -= curr_debt  # Backtrack

            return min_moves if min_moves != float('inf') else moves

        return dfs(0, debts, 0)

• Line 4-7: Build net balances (e.g., {0:-10, 1:0, 2:10}).
• Line 10-12: Filter debts (e.g., [-10, 10]).
• Line 15-29: DFS:
• Skip zeros, base case if all processed.
• For each debt, pair with opposite signs, recurse, backtrack.
• Track min moves.
• Line 31: Start DFS.
• Time Complexity: O(n * 2^n)—exponential but pruned.
• Space Complexity: O(n)—recursion stack.

It’s like a debt-clearing mastermind!

Alternative Solution: Brute Force Recursion

Why an Alternative Approach?

The brute force recursion tries all transaction permutations—O(n!) time, O(n) space—without optimization, exploring every possible settling sequence. It’s slow but intuitive, like manually testing every handshake combo. Good for small inputs or learning!

How It Works

• Step 1: Compute net balances.
• Step 2: Recurse, trying all pairings.
• Step 3: Return min transactions.

Step-by-Step Example

Example: transactions = [[0,1,10], [2,0,5]]

• Balances: {0:-5, 1:10, 2:-5}.
• Debts: [-5, 10, -5].
• Brute Force:
• Try -5 and 10: [-5], 1 move.
• Try -5 and -5: impossible, 2 moves total.
• Result: 2.

Code for Brute Force

Copy codeclass Solution:
    def minTransfers(self, transactions: List[List[int]]) -> int:
        balance = {}
        for from_id, to_id, amount in transactions:
            balance[from_id] = balance.get(from_id, 0) - amount
            balance[to_id] = balance.get(to_id, 0) + amount

        debts = [val for val in balance.values() if val != 0]
        if not debts:
            return 0

        def brute_force(debts):
            if not debts:
                return 0
            min_moves = float('inf')
            for i in range(len(debts)):
                for j in range(i + 1, len(debts)):
                    if debts[i] * debts[j] < 0:
                        new_debts = debts[:i] + debts[i+1:j] + debts[j+1:]
                        new_debts.append(debts[i] + debts[j])
                        min_moves = min(min_moves, 1 + brute_force(new_debts))
            return min_moves if min_moves != float('inf') else 0

        return brute_force(debts)

• Time Complexity: O(n!)—factorial permutations.
• Space Complexity: O(n)—recursion stack.

It’s a slow debt settler!

Comparing the Two Solutions

• DFS with Pruning (Best):
• Pros: O(n * 2^n), faster, scalable.
• Cons: O(n) space.
• Brute Force (Alternative):
• Pros: O(n!), simple.
• Cons: Impractical for n > 6.

DFS wins for efficiency.

Edge Cases and Examples

• Input: [] → 0.
• Input: [[0,1,1]] → 1.
• Input: [[0,1,10],[1,2,10]] → 2.

DFS handles all well.

Complexity Recap

• DFS: Time O(n * 2^n), Space O(n).
• Brute Force: Time O(n!), Space O(n).

DFS’s the champ.

Key Takeaways

• DFS: Prune for speed.
• Brute Force: Try all combos.
• Python Tip: Recursion optimizes—see [Python Basics](/python/basics).

Final Thoughts: Balance Those Books

LeetCode 465: Optimal Account Balancing in Python is a financial optimization thriller. DFS with pruning is your fast ledger, while brute force is a slow tally. Want more optimization? Try LeetCode 322: Coin Change or LeetCode 416: Partition Equal Subset Sum. Ready to settle some debts? Head to Solve LeetCode 465 on LeetCode (if unlocked) and balance it today—your coding skills are transaction-ready!