Optimal Account Balancing
Created: March 23, 2020 by [lek-tin]
Last updated: March 23, 2020
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 thatx ≠ yandz > 0. - Person’s IDs may not be linear, e.g. we could have the persons
0, 1, 2or we could also have the persons0, 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.
Solution
import math
class Solution:
def minTransfers(self, transactions: List[List[int]]) -> int:
self.res = math.inf
# when debts are settled, in + out = 0
accounts = collections.defaultdict(int)
for transaction in transactions:
accounts[transaction[0]] += transaction[2]
accounts[transaction[1]] -= transaction[2]
print(accounts)
# generate debts if account is non-zero
debts = [account for account in accounts.values() if account]
print(debts)
# dfs for the minimal # of transactions
self.helper(debts, 0, 0)
return self.res
def helper(self, debts, start, cnt):
n = len(debts)
print(start, n)
while start < n and debts[start] == 0:
start += 1
# everyone has been checked
if start == n:
self.res = min(self.res, cnt)
return
# check all remaining people
for i in range(start+1, n):
# when i owes/start owns or i owns/start owes
if (debts[i] < 0 and debts[start] > 0) or (debts[i] > 0 and debts[start] < 0):
# include a transaction
debts[i] += debts[start]
# count++ because
self.helper(debts, start + 1, cnt + 1)
# undo the previous transaction, continue dfs
debts[i] -= debts[start]