[LeetCode]Optimal Account Balancing

题目描述：

LeetCode 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:
2Explanation:
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:
1Explanation:
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.

题目大意：

一群朋友去度假，有时会互相借钱。例如，Alice为Bill的午餐买单，花费$10。然后Chris给Alice $5打车费。我们可以将每一笔交易模型化为一个三元组 (x, y, z)，意思是x给y花费$z。假设Alice, Bill和Chris分别标号0,1,2，以上交易可以表示为[[0, 1, 10], [2, 0, 5]]。

给定一群人的交易列表，返回结清债务关系的最小交易数。

注意：

交易以元组(x, y, z)的形式给出。x ≠ y 并且 z > 0

人员的ID不一定是线性的，例如，人员可以是0, 1, 2，也可以是0, 2, 6。

解题思路：

解法I 穷举（Brute Force）

统计每个人借出/借入的金钱总数

将借出金钱的人放入集合rich，借入金钱的人放入集合poor

问题转化为计算从rich到poor的最小“债务连线”数

尝试用rich中的每个金额与poor中的每个金额做匹配

若存在差值，则将差值加入相应集合继续搜索

这道题目似乎是NP Hard

参考资料：http://www.mathmeth.com/tom/files/settling-debts.pdf

Python代码：

class Solution(object):
    def minTransfers(self, transactions):
        """
        :type transactions: List[List[int]]
        :rtype: int
        """
        def solve(rich, poor):
            rlen, plen = len(rich), len(poor)
            if min(rlen, plen) <= 1:
                return max(rlen, plen)
            ans = 0x7FFFFFFF
            for ri in range(rlen):
                for pi in range(plen):
                    nrich = rich[:ri] + rich[ri+1:]
                    npoor = poor[:pi] + poor[pi+1:]
                    if rich[ri] < poor[pi]:
                        npoor.append(poor[pi] - rich[ri])
                    elif rich[ri] > poor[pi]:
                        nrich.append(rich[ri] - poor[pi])
                    ans = min(solve(nrich, npoor) + 1, ans)
            return ans

        loan = collections.defaultdict(int)
        for s, t, v in transactions:
            loan[s] += v
            loan[t] -= v
        rich = [v for k, v in loan.iteritems() if v > 0]
        poor = [-v for k, v in loan.iteritems() if v < 0]
        return solve(rich, poor)

解法II 记忆化搜索

与穷举法思路相同，通过保存中间计算结果可以减少重复搜索

Python代码：

class Solution(object):
    def minTransfers(self, transactions):
        """
        :type transactions: List[List[int]]
        :rtype: int
        """
        vdict = collections.defaultdict(dict)

        def solve(rich, poor):
            rlen, plen = len(rich), len(poor)
            if min(rlen, plen) <= 1:
                return max(rlen, plen)
            ri = pi = 0
            rich.sort()
            poor.sort()
            trich, tpoor = tuple(rich), tuple(poor)
            ans = vdict[trich].get(tpoor)
            if ans is not None:
                return ans
            ans = 0x7FFFFFFF
            for ri in range(rlen):
                for pi in range(plen):
                    nrich = rich[:ri] + rich[ri+1:]
                    npoor = poor[:pi] + poor[pi+1:]
                    if rich[ri] < poor[pi]:
                        npoor.append(poor[pi] - rich[ri])
                    elif rich[ri] > poor[pi]:
                        nrich.append(rich[ri] - poor[pi])
                ans = min(solve(nrich, npoor) + 1, ans)
            vdict[trich][tpoor] = ans
            return ans

        loan = collections.defaultdict(int)
        for s, t, v in transactions:
            loan[s] += v
            loan[t] -= v
        rich = [v for k, v in loan.iteritems() if v > 0]
        poor = [-v for k, v in loan.iteritems() if v < 0]
        return solve(rich, poor)