## Sunday, July 23, 2017

### [Leetcode] Optimal Account Balancing, Solution

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.
```

[Thoughts]

1. 最后的负债一定可以清零
2. 题目不要求保留原有付款关系。

[Code]
``````1:    int minTransfers(vector<vector<int>>& trans) {
2:      unordered_map<int, long> bal; // balance on each person
3:      for(auto t: trans) {
4:        bal[t] -= t;
5:        bal[t] += t;
6:      }
7:
8:      vector<long> debt;
9:      for(auto b: bal) {
10:        // only track the person who has debt
11:        if(b.second) debt.push_back(b.second);
12:      }
13:      return dfs(0, 0, debt);
14:    }
15:
16:    // get the min number of transactions starting from s
17:    int dfs(int s, int cnt, vector<long>& debt) {
18:         while (s < debt.size() && !debt[s]) ++s; // skip all zero debt
19:
20:         int res = INT_MAX;
21:         for (long i = s+1, prev = 0; i < debt.size(); ++i) {
22:        // skip same value or same sign debt
23:        if (debt[i] != prev && debt[i]*debt[s] < 0){
24:             debt[i] += debt[s];
25:          res = min(res, dfs(s+1,cnt+1, debt));
26:          debt[i]-=debt[s];
27:          prev = debt[i];
28:        }
29:      }
30:         return res < INT_MAX? res : cnt;
31:    }
``````