Hard
Optimal Account Balancing — C++
Full explanation · Time O(n * 2^n) · Space O(2^n)
// Time: O(n * 2^n), n is the size of debts.
// Space: O(2^n)
class Solution {
public:
int minTransfers(vector<vector<int>>& transactions) {
unordered_map<int, int> account;
for (const auto& transaction : transactions) {
account[transaction[0]] += transaction[2];
account[transaction[1]] -= transaction[2];
}
vector<int> debts;
for (const auto& [_, debt] : account) {
if (debt) {
debts.emplace_back(debt);
}
}
vector<int> dp(1 << debts.size());
vector<int> sums(1 << debts.size());
for (int i = 0; i < dp.size(); ++i) {
for (int j = 0, bit = 1; j < debts.size(); ++j, bit <<= 1) {
if ((i & bit) == 0) {
auto nxt = i | bit;
sums[nxt] = sums[i] + debts[j];
if (sums[nxt] == 0) {
dp[nxt] = max(dp[nxt], dp[i] + 1);
} else {
dp[nxt] = max(dp[nxt], dp[i]);
}
}
}
}
return debts.size() - dp.back();
}
};