[LeetCode] 465. Optimal Account Balancing 最優賬戶平衡

 

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.

 

這道題給了一堆某人欠某人多少錢這樣的賬單，問經過優化后最少還剩幾個。其實就相當於一堆人出去玩，某些人可能幫另一些人墊付過花費，最后結算總花費的時候可能你欠着別人的錢，其他人可能也欠你的欠，需要找出簡單的方法把所有欠賬都還清就行了。這道題的思路跟之前那道 Evaluate Division 有些像，都需要對一組數據顛倒順序處理。這里使用一個 HashMap 來建立每個人和其賬戶的映射，其中賬戶若為正數，說明其他人欠你錢；如果賬戶為負數，說明你欠別人錢。對於每份賬單，前面的人就在 HashMap 中減去錢數，后面的人在哈希表中加上錢數。這樣每個人就都有一個賬戶了，接下來要做的就是合並賬戶，看最少需要多少次匯款。先統計出賬戶值不為0的人數，因為如果為0了，表明你既不欠別人錢，別人也不欠你錢，如果不為0，把錢數放入一個數組 accnt 中，然后調用遞歸函數。在遞歸函數中，首先跳過為0的賬戶，然后看若此時 start 已經是 accnt 數組的長度了，說明所有的賬戶已經檢測完了，用 cnt 來更新結果 res。否則就開始遍歷之后的賬戶，如果當前賬戶和之前賬戶的錢數正負不同的話，將前一個賬戶的錢數加到當前賬戶上，這很好理解，比如前一個賬戶錢數是 -5，表示張三欠了別人5塊錢，當前賬戶錢數是5，表示某人欠了李四5塊錢，那么張三給李四5塊，這兩人的賬戶就都清零了。然后調用遞歸函數，此時從當前改變過的賬戶開始找，cnt 表示當前的轉賬數，需要加1，后面別忘了復原當前賬戶的值，典型的遞歸寫法，參見代碼如下：

 

class Solution {
public:
    int minTransfers(vector<vector<int>>& transactions) {
        int res = INT_MAX;
        unordered_map<int, int> m;
        for (auto t : transactions) {
            m[t[0]] -= t[2];
            m[t[1]] += t[2];
        }
        vector<int> accnt;
        for (auto a : m) {
            if (a.second != 0) accnt.push_back(a.second);
        }
        helper(accnt, 0, 0, res);
        return res;
    }
    void helper(vector<int>& accnt, int start, int cnt, int& res) {
        int n = accnt.size();
        while (start < n && accnt[start] == 0) ++start;
        if (start == n) {
            res = min(res, cnt);
            return;
        }
        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];
                helper(accnt, start + 1, cnt + 1, res);
                accnt[i] -= accnt[start];
            }
        }
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/465

 

類似題目：

Evaluate Division

 

參考資料：

https://leetcode.com/problems/optimal-account-balancing/

https://leetcode.com/problems/optimal-account-balancing/discuss/95369/share-my-on-npc-solution-tle-for-large-case

https://leetcode.com/problems/optimal-account-balancing/discuss/95355/11-liner-9ms-DFS-solution-(detailed-explanation)

 

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