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.
  • Brutal force DFS

public int minTransfers(int[][] transactions) {
    HashMap<Integer, Integer> map = new HashMap<>();
    for(int[] t : transactions) {
        map.put(t[0], map.getOrDefault(t[0],0) - t[2]);
        map.put(t[1], map.getOrDefault(t[1],0) + t[2]);
    }
    ArrayList<Integer> debt = new ArrayList<>(map.values());
    return settle(0, debt);
}

public int settle(int start, ArrayList<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(i) * debt.get(start) < 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;
}