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.

// DFS
void helper(vector<int>& account, int start, int cnt, int& res) {
    int n = account.size();
    while (start < n && account[start] == 0) ++start;
    if (start == n) {
        res = min(res, cnt);
        return;
    }
    for (int i = start + 1; i < n; ++i) {
        if (account[i] * account[start] < 0) {
            account[i] += account[start];
            helper(account, start + 1, cnt + 1, res);
            account[i] -= account[start];
        }
    }
}
int minTransfers(vector<vector<int>>& transactions) {
    unordered_map<int, int> balance;
    int res = numeric_limits<int>::max();
    for (const vector<int>& t : transactions) {
        balance[t[0]] -= t[2]; // lend
        balance[t[1]] += t[2]; // borrow
    }
    vector<int> account;
    for (auto& e : balance) {
        if (e.second != 0) account.push_back(e.second);
    }
    helper(account, 0, 0, res);
    return res;
}

// Dynamic Programming
// https://leetcode.com/problems/optimal-account-balancing/discuss/219187/Short-O(N-*-2N)-DP-solution-with-detailed-explanation-and-complexity-analysis
int minTransfers(vector<vector<int>>& transactions) {
    unordered_map<int, int> balance; // person_id -> balance
    for (auto& t: transactions) {
        balance[t[0]] -= t[2]; // lend
        balance[t[1]] += t[2]; // borrow
    }
    vector<int> account; // all non-zero balances
    for (auto& [_, b]: balance) {
        if (b) account.push_back(b);
    }
    const int n = account.size();
    vector<int> dp(1 << n); // max zero sum groups
    vector<long> sums(1 << n);
    for (int bitmask = 1; bitmask < (1 << n); ++bitmask) {
        for (int i = 0; i < n; ++i) {
            if (bitmask & (1 << i)) continue;
            int next = bitmask | (1 << i); // set bit
            sums[next] += account[i];
            if (sums[bitmask] == 0) dp[next] = max(dp[next], dp[bitmask] + 1);
            else dp[next] = max(dp[bitmask], dp[next]);
        }
    }
    return n - dp.back();
}