Merge pull request #163 from BrianLusina/feat/algorithms-backtracking

feat(algorithms, backtracking): optimal account balancing

Author: BrianLusina

DIRECTORY.md

@@ -27,6 +27,8 @@
       * [Test Decode Message](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/backtracking/decode_message/test_decode_message.py)
     * Letter Combination
       * [Test Letter Combination](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/backtracking/letter_combination/test_letter_combination.py)
+    * Optimal Account Balancing
+      * [Test Optimal Account Balancing](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/backtracking/optimal_account_balancing/test_optimal_account_balancing.py)
     * Partition String
       * [Test Partition String](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/backtracking/partition_string/test_partition_string.py)
     * Permutations

algorithms/backtracking/optimal_account_balancing/README.md

@@ -0,0 +1,76 @@
+# Optimal Account Balancing
+
+Given a list of transactions, where each transaction is represented as transactions[i]=[fromi, toi, amounti], indicating
+that the person fromi gave amounti to the person toi.
+
+Return the minimum number of transactions needed to settle all debts.
+
+## Constraints
+
+- 1 ≤ transactions.length ≤ 10
+- transactions[i].length == 3
+- 0 ≤ fromi, toi ≤ 10
+- 1 ≤ amounti ≤ 100 
+- fromi ≠ toi
+
+## Examples
+
+![Example 1](./images/examples/optimal_account_balancing_example_1.png)
+![Example 2](./images/examples/optimal_account_balancing_example_2.png)
+![Example 3](./images/examples/optimal_account_balancing_example_3.png)
+
+## Solution
+
+This solution calculates the minimum number of transactions required to settle debts among a group of people based on an
+initial set of transactions. It begins by computing each person’s net balance, indicating how much they owe or are owed.
+Any individuals with a zero net balance are ignored, as they have no outstanding debts or credits.
+
+Once the net balances are calculated, the algorithm uses a recursive approach to settle the remaining balances. It pairs
+individuals with opposite balances (i.e., those who owe with those who are owed) to reduce them to zero with the minimum
+number of transactions. The solution explores various pairings and tracks each combination’s cost (number of transactions),
+ensuring optimal results. If a particular path doesn’t minimize the transactions, it backtracks to explore alternative
+pairings.
+
+Here’s the step-by-step implementation of the solution:
+
+- For each transaction, decrease the balance of the person who gave the money and increase the balance of the person who 
+  received it.
+
+- Ignore people with a zero balance after all transactions, as they neither owe nor are owed.
+- Use depth-first search (DFS) to recursively calculate the minimum number of transactions required to settle all balances. 
+  - Base case: If the current person reaches n (number of people with non-zero balances), meaning all balances are zero
+    or settled, return 0.
+  - For each next person:
+    - If the current and next person have opposite sign balance:
+      - Temporarily add the current person’s balance to the next person’s balance. 
+      - Recursively call DFS with the next person’s index. 
+      - Track the minimum of the existing cost and the new cost calculated by DFS. 
+      - **Backtrack**: Restore the balance to its original state by reversing the temporary addition.
+  - Return the minimum cost for all possible transaction paths.
+
+Let’s look at the following illustration to get a better understanding of the solution:
+
+![Solution 1](./images/solutions/optimal_account_balancing_solution_1.png)
+![Solution 2](./images/solutions/optimal_account_balancing_solution_2.png)
+![Solution 3](./images/solutions/optimal_account_balancing_solution_3.png)
+![Solution 4](./images/solutions/optimal_account_balancing_solution_4.png)
+![Solution 5](./images/solutions/optimal_account_balancing_solution_5.png)
+![Solution 6](./images/solutions/optimal_account_balancing_solution_6.png)
+![Solution 7](./images/solutions/optimal_account_balancing_solution_7.png)
+![Solution 8](./images/solutions/optimal_account_balancing_solution_8.png)
+![Solution 9](./images/solutions/optimal_account_balancing_solution_9.png)
+![Solution 10](./images/solutions/optimal_account_balancing_solution_10.png)
+![Solution 11](./images/solutions/optimal_account_balancing_solution_11.png)
+![Solution 12](./images/solutions/optimal_account_balancing_solution_12.png)
+![Solution 13](./images/solutions/optimal_account_balancing_solution_13.png)
+
+### Time Complexity
+
+The time complexity is O((n−1)!), where n is the number of persons. This complexity arises because, in the initial call
+to dfs(0), there are n−1 possible choices for the next person, each leading to a recursive call to dfs(1), and so on.
+This results in a chain of recursive calls multiplied by the choices at each level, giving a factorial pattern (n−1)!.
+
+### Space Complexity
+
+The space complexity is O(n), where n is the number of unique amounts. This is because both balance_map and balance can
+hold at most n amounts.

algorithms/backtracking/optimal_account_balancing/__init__.py

@@ -0,0 +1,91 @@
+from typing import List, DefaultDict
+from collections import defaultdict
+
+
+def min_transfers_dfs(transactions: List[List[int]]) -> int:
+    if not transactions:
+        return 0
+
+    # Net balances where the key is the person and the value is their balance
+    net_balances: DefaultDict[int, int] = defaultdict(int)
+
+    # Populate net balances
+    for transaction in transactions:
+        sender, receiver, amount = transaction
+        net_balances[sender] -= amount
+        net_balances[receiver] += amount
+
+    # Remove zero balances
+    non_zero_balances = [amt for amt in net_balances.values() if amt != 0]
+    number_of_non_zero_balances = len(non_zero_balances)
+
+    if non_zero_balances == 0:
+        return 0
+
+    def dfs(current: int, balances: List[int]) -> int:
+        # Skip settled accounts, move to the next person with non-zero balaance
+        while current < number_of_non_zero_balances and balances[current] == 0:
+            current += 1
+        # All accounts settled
+        if current >= number_of_non_zero_balances:
+            return 0
+
+        min_transactions = float("inf")
+        # Try to settle non_zero_balances[start] by paring with another opposite sign balance
+        for j in range(current + 1, number_of_non_zero_balances):
+            # One owes, the other is owed
+            if balances[current] * balances[j] < 0:
+                balances[j] += balances[current]
+                # recurse for remaining transactions after settling this one
+                min_transactions = min(min_transactions, 1 + dfs(current + 1, balances))
+                # backtrack, undo the settlement
+                non_zero_balances[j] -= non_zero_balances[current]
+
+        return min_transactions
+
+    return dfs(0, non_zero_balances)
+
+
+def min_transfers_backtrack(transactions: List[List[int]]) -> int:
+    if not transactions:
+        return 0
+
+    # Net balances where the key is the person and the value is their balance
+    net_balances: DefaultDict[int, int] = defaultdict(int)
+
+    # Populate net balances
+    for transaction in transactions:
+        sender, receiver, amount = transaction
+        net_balances[sender] -= amount
+        net_balances[receiver] += amount
+
+    # Remove zero balances
+    non_zero_balances = [amt for amt in net_balances.values() if amt != 0]
+    number_of_non_zero_balances = len(non_zero_balances)
+
+    if non_zero_balances == 0:
+        return 0
+
+    def backtrack(start: int) -> int:
+        # Skip settled accounts
+        while start < number_of_non_zero_balances and non_zero_balances[start] == 0:
+            start += 1
+        # All accounts settled
+        if start >= number_of_non_zero_balances:
+            return 0
+
+        min_transactions = float("inf")
+        # Try to settle non_zero_balances[start] by paring with another opposite sign balance
+        for j in range(start + 1, number_of_non_zero_balances):
+            # One owes, the other is owed
+            if non_zero_balances[start] * non_zero_balances[j] < 0:
+                non_zero_balances[j] += non_zero_balances[start]
+                # recurse for remaining transactions after settling this one
+                min_transactions = min(min_transactions, 1 + backtrack(start + 1))
+                # backtrack, undo the settlement
+                non_zero_balances[j] -= non_zero_balances[start]
+                if non_zero_balances[j] + non_zero_balances[start] == 0:
+                    break
+        return min_transactions
+
+    return backtrack(0)