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, 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 thatx ≠ yandz > 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.

S： recursion/backtracking O(n!) fractorial

用debt记录每个人的负债情况，得到负债数组amount。对amount从0开始的每个负债数amount[start]，找到所有跟该负债符号相反的负债amount[i]，将amount[start]付给amount[i](相加). amount[start]清零，可得到0-start的负债全部清零。

dfs直到start == amount.size()即全部负债清零

    int minTransfers(vector<vector<int>>& transactions) {
        int n = transactions.size();
        if (n < 2) return n;
        unordered_map<int, int> debt;
        for (vector<int>& txn : transactions) {
            debt[txn[0]] -= txn[2];
            debt[txn[1]] += txn[2];
        }
        vector<int> amount;
        for (const auto& d : debt) {
            amount.push_back(d.second);
        }
        int result = INT_MAX;
        dfs(amount, 0, 0, result);
        return result;
    }

    void dfs(vector<int>& amount, int start, int count, int& result) {
        while (start < amount.size() && amount[start] == 0) start++; //debt from 0 to start is clean
        if (start >= amount.size()) {
            result = min(result, count);
            return;
        }     
        for (int i = start + 1; i < amount.size(); ++i) {
            if (amount[i] != 0 && ((amount[start] > 0) ^ (amount[i] > 0))) {                
                amount[i] += amount[start];
                dfs(amount, start + 1, count + 1, result);
                amount[i] -= amount[start];                 
            }
        }        
    }