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 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.

 

这道题给了一堆某人欠某人多少钱这样的账单,问经过优化后最少还剩几个。其实就相当于一堆人出去玩,某些人可能帮另一些人垫付过花费,最后结算总花费的时候可能你欠着别人的钱,其他人可能也欠你的欠,需要找出简单的方法把所有欠账都还清就行了。这道题的思路跟之前那道 Evaluate Division 有些像,都需要对一组数据颠倒顺序处理。这里使用一个 HashMap 来建立每个人和其账户的映射,其中账户若为正数,说明其他人欠你钱;如果账户为负数,说明你欠别人钱。对于每份账单,前面的人就在 HashMap 中减去钱数,后面的人在哈希表中加上钱数。这样每个人就都有一个账户了,接下来要做的就是合并账户,看最少需要多少次汇款。先统计出账户值不为0的人数,因为如果为0了,表明你既不欠别人钱,别人也不欠你钱,如果不为0,把钱数放入一个数组 accnt 中,然后调用递归函数。在递归函数中,首先跳过为0的账户,然后看若此时 start 已经是 accnt 数组的长度了,说明所有的账户已经检测完了,用 cnt 来更新结果 res。否则就开始遍历之后的账户,如果当前账户和之前账户的钱数正负不同的话,将前一个账户的钱数加到当前账户上,这很好理解,比如前一个账户钱数是 -5,表示张三欠了别人5块钱,当前账户钱数是5,表示某人欠了李四5块钱,那么张三给李四5块,这两人的账户就都清零了。然后调用递归函数,此时从当前改变过的账户开始找,cnt 表示当前的转账数,需要加1,后面别忘了复原当前账户的值,典型的递归写法,参见代码如下:

 

class Solution {
public:
    int minTransfers(vector<vector<int>>& transactions) {
        int res = INT_MAX;
        unordered_map<int, int> m;
        for (auto t : transactions) {
            m[t[0]] -= t[2];
            m[t[1]] += t[2];
        }
        vector<int> accnt;
        for (auto a : m) {
            if (a.second != 0) accnt.push_back(a.second);
        }
        helper(accnt, 0, 0, res);
        return res;
    }
    void helper(vector<int>& accnt, int start, int cnt, int& res) {
        int n = accnt.size();
        while (start < n && accnt[start] == 0) ++start;
        if (start == n) {
            res = min(res, cnt);
            return;
        }
        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];
                helper(accnt, start + 1, cnt + 1, res);
                accnt[i] -= accnt[start];
            }
        }
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/465

 

类似题目:

Evaluate Division

 

参考资料:

https://leetcode.com/problems/optimal-account-balancing/

https://leetcode.com/problems/optimal-account-balancing/discuss/95369/share-my-on-npc-solution-tle-for-large-case

https://leetcode.com/problems/optimal-account-balancing/discuss/95355/11-liner-9ms-DFS-solution-(detailed-explanation)

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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