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.

思路：首先我们需要知道最后每个人到底是欠被人还是别人欠自己，所以需要又一个map来保存最后的值balance。然后我们把balance不为0的放在另一个数组里，然后对这个数组进行dfs处理，看如何把数组里的值进行分组，
使得他们的和为0并且次数最少。
在实现dfs的时候，下面的方法比较tricky，更一般的方法（但是空间复杂度会升高很多）是把数组中两个数（正数和负数）取出来，把非0的和以及原数组里剩下的和放在另一个数组里，进行下一次递归，每次递归的时候，把
transaction的次数加一就可以了。因为是dfs，所以我们需要找到一个global最小的即可。

class Solution {
    public int minTransfers(int[][] transactions) {
        Map<Integer, Long> map = new HashMap<>();
        for (int[] transaction : transactions) {
            long balance1 = map.getOrDefault(transaction[0], 0L);
            long balance2 = map.getOrDefault(transaction[1], 0L);
            map.put(transaction[0], balance1 - transaction[2]);
            map.put(transaction[1], balance2 + transaction[2]);
        }

        List<Long> list = new ArrayList<>();
        for (long val : map.values()) {
            if (val != 0) {
                list.add(val);
            }
        }
        int matchCount = removeMatch(list);
        return matchCount + helper(list, 0);
    }
    //pre-process to decrease dfs processing time, this step can be optional.
    private int removeMatch(List<Long> list) {
        Collections.sort(list);
        int left = 0, right = list.size() - 1;
        int matchCount = 0;
        while (left < right) {
            if (list.get(left) + list.get(right) == 0) {
                list.remove(left);
                list.remove(right - 1);
                right -= 2;
                matchCount++;
            } else if (list.get(left) + list.get(right) < 0) {
                left++;
            } else {
                right--;
            }
        }
        return matchCount;
    }

    private int helper(List<Long> list, int start) {
        int result = Integer.MAX_VALUE;
        int n = list.size();
        while (start < n && list.get(start) == 0) {
            start++;
        }
        if (start == n) {
            return 0;
        }
        //这个for loop来处理dfs特别的巧，他把整个数组放在下一次递归里，并没有重新创建一个sublist，会节约空间，但是面试的时候，
        // 如果面试官不知道整个方法，可能会比较麻烦，因为这种方法不是那么容易理解。
        for (int i = start + 1; i < n; i++) {
            if (list.get(i) * list.get(start) < 0) {
                list.set(i, list.get(i) + list.get(start));
                result = Math.min(result, 1 + helper(list, start + 1));
                list.set(i, list.get(i) - list.get(start));
            }
        }
        return result;
    }
}