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:

  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.

Rust Solution

struct Solution;

use std::collections::HashMap;

impl Solution {
    fn min_transfers(transactions: Vec<Vec<i32>>) -> i32 {
        let mut balance: HashMap<i32, i32> = HashMap::new();
        for t in transactions {
            *balance.entry(t[0]).or_default() -= t[2];
            *balance.entry(t[1]).or_default() += t[2];
        }
        let mut debts = vec![];
        for &v in balance.values() {
            if v != 0 {
                debts.push(v);
            }
        }
        let n = debts.len();
        let mut res = std::usize::MAX;
        Self::dfs(0, 0, &mut debts, &mut res, n);
        res as i32
    }

    fn dfs(start: usize, cur: usize, debts: &mut Vec<i32>, min: &mut usize, n: usize) {
        if start == n {
            *min = cur.min(*min);
        } else {
            if debts[start] == 0 {
                Self::dfs(start + 1, cur, debts, min, n);
            } else {
                let transaction = debts[start];
                debts[start] -= transaction;
                for i in start + 1..n {
                    if debts[i] * transaction < 0 {
                        debts[i] += transaction;
                        Self::dfs(start + 1, cur + 1, debts, min, n);
                        debts[i] -= transaction;
                    }
                }
                debts[start] += transaction;
            }
        }
    }
}

#[test]
fn test() {
    let transactions = vec_vec_i32![[0, 1, 10], [2, 0, 5]];
    let res = 2;
    assert_eq!(Solution::min_transfers(transactions), res);
    let transactions = vec_vec_i32![[0, 1, 10], [1, 0, 1], [1, 2, 5], [2, 0, 5]];
    let res = 1;
    assert_eq!(Solution::min_transfers(transactions), res);
}

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