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Leetcode: Optimal Account Balancing

2024-08-28 14:41:32   215人阅读

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:

A transaction will be given as a tuple (x, y, z). Note that x ≠ y and z > 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.
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Backtracking: time complexity O(N!)

Use HashMap to store the initial debts of each person, negative means the person sends money to others, positive means the person gets money from others.

now if the map value is 0, which means the person is all set, free of debts.

Only consider those people with debts(either positive or negative)

store them in an array, use backtracking and greedy to clear each person‘s debts from 1st person till last one.

How to clear one person‘s debt? find a person 2 in the following array that has opposite sign(+->-  or - -> +), and clear person1‘s whole debt with person2 only. 

Here‘s the trick: example: [7, -6, -1], one obvious optimal solution is person1 pay $6 to person2, and pay $1 to person3. Notice that this optimal solution is equivalent to another solution:

person1 pay $7 to person2, and person2 pay $1 to person3. So when doing DFS, everytime we only consider clearing person1‘s debt wholly with another 1 person, we don‘t need to consider clearing with other more people, cause clearing with 1 person is already guaranteed to be optimal.

 

This problem still has some debates in discussion, will check later

 1 public class Solution {
 2     int res = Integer.MAX_VALUE;
 3     public int minTransfers(int[][] transactions) {
 4         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
 5         for (int[] transaction : transactions) {
 6             map.put(transaction[0], map.getOrDefault(transaction[0], 0)-transaction[2]);
 7             map.put(transaction[1], map.getOrDefault(transaction[1], 0)+transaction[2]);
 8         }
 9         ArrayList<Integer> depts = new ArrayList<Integer>();
10         for (int dept : map.values()) {
11             if (dept != 0) depts.add(dept);
12         }
13         helper(depts, 0, 0);
14         return res;
15     }
16     
17     public void helper(ArrayList<Integer> depts, int start, int count) {
18         while (start<depts.size() && depts.get(start)==0) start++;
19         if (start == depts.size()) {
20             res = Math.min(res, count);
21             return;
22         }
23         for (int i=start+1; i<depts.size(); i++) {
24             if (depts.get(start)<0&&depts.get(i)>0 || depts.get(start)>0&&depts.get(i)<0) {
25                 depts.set(i, depts.get(i)+depts.get(start));
26                 //int store = depts.get(start);
27                 //depts.set(start, 0);
28                 helper(depts, start+1, count+1);
29                 //depts.set(start, store);
30                 depts.set(i, depts.get(i)-depts.get(start));
31             }
32         }
33     }
34 }

 

Leetcode: Optimal Account Balancing


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