Comment on page

# 465.Optimal-Account-Balancing

## 题目描述

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.

## 代码

### Approach #1 Backtracing

Time: O(N) && Space: O(N)
class Solution {
public int minTransfers(int[][] transactions) {
Map<Integer, Integer> m = new HashMap<>();
for (int[] t: transactions {
m.put(t, m.getOrDefault(t, 0) - t);
m.put(t, m.getOrDefault(t, 0) + t);
}
return settle(0, new ArrayList<>(m.values()));
}
int settle(int start, List<Integer> debt) {
while (start < debt.size() && debt.get(start) == 0) {
start++;
}
if (start == debt.size()) return 0;
int r = Integer.MAX_VALUE;
for (int i = start + 1; i < debt.size(); i++) {
if (debt.get(start) * debt.get(i) < 0) {
debt.set(i, debt.get(i) + debt.get(start));
r = Math.min(r, 1 + settle(start + 1, debt));
debt.set(i, debt.get(i) - debt.get(start));
}
}
return r;
}
}