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.

Solution:

With all the given transactions, in the end, each person with ID = id will have an overall balance bal[id]. Note that the id value or any person coincidentally with 0 balance is irrelevant to debt settling count, so we can simply use an array debt[] to store all non-zero balances, where



Starting from first debt debt[0], we look for all other debts debt[i] (i>0) which have opposite sign to debt[0]. Then each such debt[i] can make one transaction debt[i] += debt[0] to clear the person with debt debt[0]. From now on, the person with debt debt[0] is dropped out of the problem and we recursively drop persons one by one until everyone's debt is cleared meanwhile updating the minimum number of transactions during DFS.


Note: Thanks to @KircheisHe who found the following great paper about the debt settling problem:



The question can be transferred to a 3-partition problem, which is NP-Complete.

class Solution {
    public int minTransfers(int[][] transactions) {
        Map<Integer, Integer> wallets = new HashMap();
        for (int[] txn : transactions) {
            int aBalance = wallets.getOrDefault(txn[0], 0);
            int bBalance = wallets.getOrDefault(txn[1], 0);
            wallets.put(txn[0], aBalance - txn[2]);
            wallets.put(txn[1], bBalance + txn[2]);
            
        }
        // System.out.println(wallets.values());
        return backtracking(new ArrayList(wallets.values()), 0);
    }
    
    private int backtracking(List<Integer> debts, int start) {
        while (start < debts.size() && debts.get(start) == 0) {
            start ++;
        }
        if (start == debts.size()) return 0;
        int min = Integer.MAX_VALUE;
        for (int i = start + 1; i < debts.size(); i ++) {
            if (debts.get(start) *  debts.get(i) < 0) {
                debts.set(i, debts.get(i) + debts.get(start));
                min = Math.min(min, 1 + backtracking(debts, start + 1));
                debts.set(i, debts.get(i) - debts.get(start));
            }
        }
        return min;
    }
}