leetcode - 465. Optimal Account Balancing

Description

You are given an array of transactions transactions where transactions[i] = [fromi, toi, amounti] indicates that the person with ID = fromi gave amounti $ to the person with ID = toi.

Return the minimum number of transactions required to settle the debt.

Example 1:

Input: transactions = [[0,1,10],[2,0,5]]
Output: 2
Explanation:
Person #0 gave person #1 $10.
Person #2 gave person #0 $5.
Two transactions are needed. One way to settle the debt is person #1 pays person #0 and #2 $5 each.

Example 2:

Input: transactions = [[0,1,10],[1,0,1],[1,2,5],[2,0,5]]
Output: 1
Explanation:
Person #0 gave person #1 $10.
Person #1 gave person #0 $1.
Person #1 gave person #2 $5.
Person #2 gave person #0 $5.
Therefore, person #1 only need to give person #0 $4, and all debt is settled.

Constraints:

1 <= transactions.length <= 8
transactions[i].length == 3
0 <= fromi, toi < 12
fromi != toi
1 <= amounti <= 100

Solution

Solved after help.

Run a debt list, to store all the balance of people. Then backtrace, try every transaction.

Time complexity: $o(2^n)=o(2^8)$
Space complexity: $o(n)$

Code

class Solution:
    def minTransfers(self, transactions: List[List[int]]) -> int:
        people = {}
        for p1, p2, amount in transactions:
            people[p1] = people.get(p1, 0) - amount
            people[p2] = people.get(p2, 0) + amount
        debts = [value for value in people.values() if value != 0]
        def dfs(start_index: int) -> int:
            if start_index == len(debts):
                return 0
            if debts[start_index] == 0:
                return dfs(start_index + 1)
            res = float('inf')
            for i in range(start_index + 1, len(debts)):
                if debts[i] * debts[start_index] < 0:
                    debts[i] += debts[start_index]
                    res = min(res, 1 + dfs(start_index + 1))
                    debts[i] -= debts[start_index]
            return res if res != float('inf') else 0
        return dfs(0)