# 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.

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

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``````
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];
int loaner = transaction[i];
int money = transaction[i];
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(NN) Where
N is total number of people in a group

## Space Complexity

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