Optimal Account Balancing

A group of friends went on holiday and sometimes lent each other money. For example, Alice paid for Bill's lunch for $10. Then later Chris gave Alice $5 for a taxi ride. We can model each transaction as a tuple (x, y, z) which means person x gave person y $z. Assuming Alice, Bill, and Chris are person 0, 1, and 2 respectively (0, 1, 2 are the person's ID), the transactions can be represented as [[0, 1, 10], [2, 0, 5]].

Given a list of transactions between a group of people, return the minimum number of transactions required to settle the debt.

Note:

1. A transaction will be given as a tuple (x, y, z). Note that x ≠ y and z > 0.

2. Person's IDs may not be linear, e.g. we could have the persons 0, 1, 2 or we could also have the persons 0, 2, 6.

Example 1:

Input:
[[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:
[[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.

分析

每个人付出-，收到+，建立map， values做 debts list

dfs从某起点开始，loop加入后面每个debts，进行下一步dfs(start+1)，记得debts数组要恢复。debts==0忽略

等于每次清掉start的债务，丢掉start,然后继续下一轮，直到全部债务清除。

class Solution:
    def minTransfers(self, transactions: List[List[int]]) -> int:
        mm = collections.defaultdict(int)
        for a,b,m in transactions:
            mm[a] = mm.get(a,0) - m
            mm[b] = mm.get(b,0) + m
        debts = list(mm.values())
        n = len(debts)
        def dfs(pos):
            while pos < n and debts[pos] == 0:
                pos += 1
            if pos == n:
                return 0
            r = float('inf')
            for i in range(pos+1, n):
                if debts[i]*debts[pos] < 0:
                    debts[i] += debts[pos]
                    r= min(r,1+dfs(pos+1))
                    debts[i] -= debts[pos]
            return r
        return dfs(0)