This page explains Java solution to problem `Optimal Account Balancing`

using `backtracking`

algorithm.

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.

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.

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.

```
package com.vc.hard;
import java.util.*;
class OptimalAccountBalancing {
public int minTransfers(int[][] transaction) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < transaction.length; i++) {
int lender = transaction[i][0];
int loaner = transaction[i][1];
int money = transaction[i][2];
map.put(lender, map.getOrDefault(lender, 0) + money);
map.put(loaner, map.getOrDefault(loaner, 0) - money);
}
List<Integer> list = new ArrayList<>(map.values());
return solve(0, list);
}
private int solve(int start, List<Integer> debits) {
int res = Integer.MAX_VALUE;
while(start < debits.size() && debits.get(start) == 0) start++;
if(start == debits.size()) return 0;
for(int i = start + 1; i < debits.size(); i++) {
if(debits.get(i) * debits.get(start) < 0) {
debits.set(i, debits.get(i) + debits.get(start));
res = Math.min(res, solve(start + 1, debits) + 1);
debits.set(i, debits.get(i) - debits.get(start));
}
}
return res;
}
}
```

O(N

^{N}) Where

N is total number of people in a group

O(N) Where

N is total number of people in a group