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:

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.

Solution

(1) Java

class Solution {
    public int minTransfers(int[][] transactions) {
        if (transactions.length == 0 || transactions[0].length == 0) {
            return 0;
        }
        Map<Integer, Integer> persons = new HashMap<>();
        for (int[] trans : transactions) {
            persons.put(trans[0], persons.getOrDefault(trans[0], 0)-trans[2]);
            persons.put(trans[1], persons.getOrDefault(trans[1], 0)+trans[2]);
        }
        List<Integer> debts = new ArrayList<>();
        for (Integer deb : persons.values()) {
            if (deb != 0) {
                debts.add(deb);
            }
        }
        return setter(debts, 0);
    }

    private int setter(List<Integer> debts, int start) {  
        while (start < debts.size() && debts.get(start) == 0) {
            start++;
        } 
        if (start == debts.size()) {
            return 0;
        }
        int rst = Integer.MAX_VALUE-1;
        for (int i = start+1; i < debts.size(); i++) {
            if (debts.get(i)*debts.get(start) < 0) {
                debts.set(i, debts.get(i)+debts.get(start));
                rst = Math.min(rst, 1+setter(debts, start+1));
                debts.set(i, debts.get(i)-debts.get(start));
            }
        }
        return rst;
    }
}

465. Optimal Account Balancing