Optimal Account Balancing.py

'''
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:

A transaction will be given as a tuple (x, y, z). Note that x ≠ y and z > 0.
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(object):
    def minTransfers(self, transactions):
        """
        :type transactions: List[List[int]]
        :rtype: int
        """
        dic = {}
        for x, y, z in transactions:
            if x not in dic:
                dic[x] = 0
            if y not in dic:
                dic[y] = 0
            dic[x] -= z
            dic[y] += z
            
        return self.find(0, dic.values())
    
    def find(self, idx, vec):
        while idx < len(vec) and vec[idx] == 0:
            idx += 1
        if idx == len(vec):
            return 0
        
        res = float('inf')
        for i in xrange(idx+1, len(vec)):
            if vec[idx] * vec[i] >= 0:
                continue
            
            vec[i] += vec[idx]
            res = min(res, self.find(idx+1, vec)+1)
            vec[i] -= vec[idx]
        return res