LC 465. 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 (0,2 are the person‘s ID),the transactions can be represented as?[[0,10],[2,5]].

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,z). Note that?x ≠ y?and?z > 0.
2. Person‘s IDs may not be linear,e.g. we could have the persons 0,2 or we could also have the persons 0,2,6.

?

Example 1:

Input:
[[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,[1,1],5],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.


Runtime 24 ms,faster than 30.39%

这题本来以为是图，结果是数组的题。先建立每一个人的一个账户，然后DFS，DFS的时候从0-n，把第i个账户的钱都转到某一个j，这两个账户的金额是相反的，
因为只有相反的才会有交易。

class Solution {
public:
  int minTransfers(vector<vector<int>>& transactions) {
    map<int,int> m;
    for(auto t: transactions){
      m[t[0]] -= t[2];
      m[t[1]] += t[2];
    }
    vector<int> accnt(m.size());
    int cnt = 0;
    for(auto a : m){
      if(a.second != 0) accnt[cnt++] = a.second;
    }
    return helper(accnt,0,cnt,0);
  }
  int helper(vector<int>& accnt,int start,int n,int num){
    int ret = INT_MAX;
    while(start < n && accnt[start] == 0) start++;
    for(int i=start+1; i<n; i++){
      if((accnt[start] < 0 && accnt[i] > 0) || (accnt[start] > 0 && accnt[i] < 0)){
        accnt[i] += accnt[start];//加入第i个账户中，这个时候start账户已经没有钱了，可以进行下一个账户清理了
        ret = min(ret,helper(accnt,start+1,n,num+1));//DFS，保存最小值
        accnt[i] -= accnt[start];
      }
    }
    return ret == INT_MAX ? num : ret;
  }
};