Optimal Account Balancing 最优账户平衡

A group of friends went on holiday and sometimes lent each other money. For example, Alice paid for Bill's lunch for 10.ThenlaterChrisgaveAlice10.ThenlaterChrisgaveAlice5 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.

 

class Solution {
public:
    int minTransfers(vector<vector<int>>& transactions) {
        unordered_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 res = INT_MAX;
        while (start < n && accnt[start] == 0) ++start;
        for (int i = start + 1; i < n; ++i) {
            if ((accnt[i] < 0 && accnt[start] > 0) || (accnt[i] > 0 && accnt[start] < 0)) {
                accnt[i] += accnt[start];
                res = min(res, helper(accnt, start + 1, n, num + 1));
                accnt[i] -= accnt[start];
            }
        }
        return res == INT_MAX ? num : res;
    }
};