Optimal Account Balancing

This page explains Java solution to problem Optimal Account Balancing using backtracking algorithm.

Problem Statement

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: 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.

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

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;
    }
}

Time Complexity

O(N^N) Where
N is total number of people in a group

Space Complexity

O(N) Where
N is total number of people in a group