# 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.
``````

``````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

Evaluate Division

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…)

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation