465.Optimal-Account-Balancing

465. Optimal Account Balancing

题目地址

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

题目描述

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.

代码

Approach #1 Backtracing

Time: O(N) && Space: O(N)

class Solution {
  public int minTransfers(int[][] transactions) {
        Map<Integer, Integer> m = new HashMap<>();
    for (int[] t: transactions {
      m.put(t[0], m.getOrDefault(t[0], 0) - t[2]);
      m.put(t[1], m.getOrDefault(t[1], 0) + t[2]);
    }
    return settle(0, new ArrayList<>(m.values()));
  }

  int settle(int start, List<Integer> debt) {
    while (start < debt.size() && debt.get(start) == 0) {
      start++;
    }
    if (start == debt.size())        return 0;
    int r = Integer.MAX_VALUE;
    for (int i = start + 1; i < debt.size(); i++) {
      if (debt.get(start) * debt.get(i) < 0) {
        debt.set(i, debt.get(i) + debt.get(start));
        r = Math.min(r, 1 + settle(start + 1, debt));
        debt.set(i, debt.get(i) - debt.get(start));
      }
    }

    return r;
  }
}