A group of friends went on holiday and sometimes lent each other money. For example, Alice paid for Bill's lunch for 10.ThenlaterChrisgaveAlice10.ThenlaterChrisgaveAlice5 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:
- A transaction will be given as a tuple (x, y, z). Note that
x ≠ yandz > 0. - 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.
一堆人出去度假，大家互相帮别人垫付过钱，最后结算平衡总花费，可能你欠着别人的钱，其他人可能也欠你的钱，有的账就不用还了，找出最少需要多少次交易。
解法: 首先计算出每个人的最后盈亏，然后用随机的算法找到比较可能的最小值。对于所有交易，每个人最终将有一个总负债bal[id]，所有负债为0的人不需要支付交易，用一个精炼的负债表debt[]表示负债不为零的用户的负债信息，其中：debt[i] > 0 表示该人需要向其他人支付debt[i]的金钱；debt[i] < 0 表示该人需要接受来自其他人总计debt[i]的金钱。
从第一个负债debt[0]开始，看看所有它后面的人的负债信息，当搜索到第一个和debt负债信息sign相反的人i的信息时，就试图平衡这两个人之间的负债关系（即在0和i之间产生一个交易，使得debt[0]为0）。从此以后，用户0就负债清零了，所以再递归地查找后面的负债清零状况。在DFS的过程中，可以维护一个最小的交易次数，并最终返回。
谷歌题
Java:
public class Solution {
public int minTransfers(int[][] transactions) {
Map balances = new HashMap<>();
for(int[] tran: transactions) {
balances.put(tran[0], balances.getOrDefault(tran[0], 0) - tran[2]);
balances.put(tran[1], balances.getOrDefault(tran[1], 0) + tran[2]);
}
List poss = new ArrayList<>();
List negs = new ArrayList<>();
for(Map.Entry balance : balances.entrySet()) {
int val = balance.getValue();
if (val > 0) poss.add(val);
else if (val < 0) negs.add(-val);
}
int min = Integer.MAX_VALUE;
Stack ps = new Stack<>();
Stack ns = new Stack<>();
for(int i = 0; i < 10; i++) {
for(int pos: poss) {
ps.push(pos);
}
for(int neg: negs) {
ns.push(neg);
}
int count = 0;
while (!ps.isEmpty()) {
int p = ps.pop();
int n = ns.pop();
if (p > n) {
ps.push(p - n);
} else if (p < n) {
ns.push(n - p);
}
count++;
}
min = Math.min(min, count);
Collections.shuffle(poss);
Collections.shuffle(negs);
}
return min;
}
} Java:
public int minTransfers(int[][] transactions) {
if (transactions == null || transactions.length == 0 || transactions[0].length == 0)
return 0;
//calculate delta for each account
Map accountToDelta = new HashMap();
for (int[] transaction : transactions) {
int from = transaction[0];
int to = transaction[1];
int val = transaction[2];
if (!accountToDelta.containsKey(from)) {
accountToDelta.put(from, 0);
}
if (!accountToDelta.containsKey(to)) {
accountToDelta.put(to, 0);
}
accountToDelta.put(from, accountToDelta.get(from) - val);
accountToDelta.put(to, accountToDelta.get(to) + val);
}
List deltas = new ArrayList();
for (int delta : accountToDelta.values()) {
if (delta != 0)
deltas.add(delta);
}
//since minTransStartsFrom is slow, we can remove matched deltas to optimize it
//for example, if account A has balance 5 and account B has balance -5, we know
//that one transaction from B to A is optimal.
int matchCount = removeMatchDeltas(deltas);
//try out all possibilities to get minimum number of transactions
return matchCount + minTransStartsFrom(deltas, 0);
}
private int removeMatchDeltas(List deltas) {
Collections.sort(deltas);
int left = 0;
int right = deltas.size() - 1;
int matchCount = 0;
while (left < right) {
if (-1 * deltas.get(left) == deltas.get(right)) {
deltas.remove(left);
deltas.remove(right - 1);
right -= 2;
matchCount++;
} else if (-1 * deltas.get(left) > deltas.get(right)) {
left++;
} else {
right--;
}
}
return matchCount;
}
private int minTransStartsFrom(List deltas, int start) {
int result = Integer.MAX_VALUE;
int n = deltas.size();
while (start < n && deltas.get(start) == 0)
start++;
if (start == n)
return 0;
for (int i = start + 1; i < n; i++) {
if ((long) deltas.get(i) * deltas.get(start) < 0) {
deltas.set(i, deltas.get(i) + deltas.get(start));
result = Math.min(result, 1 + minTransStartsFrom(deltas, start + 1));
deltas.set(i, deltas.get(i) - deltas.get(start));
}
}
return result;
} Python:
# Time: O(n * 2^n), n is the size of the debt.
# Space: O(n * 2^n)
import collections
class Solution(object):
def minTransfers(self, transactions):
"""
:type transactions: List[List[int]]
:rtype: int
"""
account = collections.defaultdict(int)
for transaction in transactions:
account[transaction[0]] += transaction[2]
account[transaction[1]] -= transaction[2]
debt = []
for v in account.values():
if v:
debt.append(v)
if not debt:
return 0
n = 1 << len(debt)
dp, subset = [float("inf")] * n, []
for i in xrange(1, n):
net_debt, number = 0, 0
for j in xrange(len(debt)):
if i & 1 << j:
net_debt += debt[j]
number += 1
if net_debt == 0:
dp[i] = number - 1
for s in subset:
if (i & s) == s:
dp[i] = min(dp[i], dp[s] + dp[i - s])
subset.append(i)
return dp[-1] C++:
class Solution {
public:
int minTransfers(vector>& transactions) {
unordered_map m;
for (auto t : transactions) {
m[t[0]] -= t[2];
m[t[1]] += t[2];
}
vector accnt(m.size());
int cnt = 0;
for (auto a : m) {
if (a.second != 0) accnt[cnt++] = a.second;
}
return helper(accnt, 0, cnt, 0);
}
int helper(vector& accnt, int start, int n, int num) {
int res = INT_MAX;
while (start < n && accnt[start] == 0) ++start;
for (int i = start + 1; i < n; ++i) {
if ((accnt[i] < 0 && accnt[start] > 0) || (accnt[i] > 0 && accnt[start] < 0)) {
accnt[i] += accnt[start];
res = min(res, helper(accnt, start + 1, n, num + 1));
accnt[i] -= accnt[start];
}
}
return res == INT_MAX ? num : res;
}
}; C++:
class Solution {
public:
int minTransfers(vector>& transactions) {
unordered_map bal; // each person's overall balance
for(auto &t: transactions) {
bal[t[0]] -= t[2];
bal[t[1]] += t[2];
}
for(auto &p: bal) {
if(p.second) {
debt.push_back(p.second);
}
}
return dfs(0, 0);
}
private:
int dfs(int s, int cnt) { // min number of transactions to settle starting from debt[s]
while (s < debt.size() && !debt[s]) {
++s; // get next non-zero debt
}
int res = INT_MAX;
for (long i = s + 1, prev = 0; i < debt.size(); ++i) {
if (debt[i] != prev && debt[i] * debt[s] < 0) { // skip already tested or same sign debt
debt[i] += debt[s];
res = min(res, dfs(s + 1,cnt + 1)); // backtracking
debt[i] -= debt[s];
prev = debt[i];
}
}
return res < INT_MAX? res : cnt;
}
vector debt; // all non-zero balances
};
C++:
class Solution {
public:
int minTransfers(vector>& transactions) {
unordered_map mp;
for (auto x : transactions) {
mp[x[0]] -= x[2];
mp[x[1]] += x[2];
}
vector in;
vector out;
for (auto x : mp) {
if (x.second < 0) out.push_back(-x.second);
else if (x.second > 0) in.push_back(x.second);
}
int amount = 0;
for (auto x : in) amount += x;
if (amount == 0) return 0;
int res = (int)in.size() + (int)out.size() - 1;
dfs(in, out, 0, 0, amount, 0, res);
return res;
}
void dfs(vector &in, vector &out, int i, int k,
int amount, int step, int &res) {
if (step >= res) return;
if (amount == 0) {
res = step;
return;
}
if (in[i] == 0) {
++i;
k = 0;
}
for (int j = k; j < out.size(); j++) {
int dec = min(in[i], out[j]);
if (dec == 0) continue;
in[i] -= dec;
out[j] -= dec;
dfs(in, out, i, j + 1, amount - dec, step + 1, res);
in[i] += dec;
out[j] += dec;
}
}
};