# 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)
                    
        
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://nataliekung.gitbook.io/ladder_code/karat/optimal-account-balancing.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.