[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]
很有意思的一道题。首先，对图进行处理，算清楚每个人的负债， 有两个前提：
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。

所以，对图做个dfs即可。

[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[0]] -= t[2];  
5:        bal[t[1]] += t[2];  
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:    }