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.
一刷
题解：
提供一系列的transaction, 问还需要几次transaction能平衡balance
最简单的方法：backtracking
从pos开始，如果有debts[i]与debts[pos]符号不同，那么进行交易。
class Solution {
// DFS
public int minTransfers(int[][] transactions) {
Map<Integer, Long> map = new HashMap<>();
for(int[] t : transactions){
long val1 = map.getOrDefault(t[0], 0L);//balance
long val2 = map.getOrDefault(t[1], 0L);
map.put(t[0], val1-t[2]);
map.put(t[1], val2+t[2]);
}
List<Long> list = new ArrayList<>();
for(long val : map.values()){
if(val!=0) list.add(val);
}
Long[] debts = new Long[list.size()];
debts = list.toArray(debts);
return helper(debts, 0, 0);
}
int helper(Long[] debts, int pos, int count ){
while(pos<debts.length && debts[pos] == 0) pos++;
if(pos>=debts.length) return count;
int res = Integer.MAX_VALUE;
for(int i=pos+1; i<debts.length; i++){
if((debts[pos] ^ debts[i])<0){
debts[i] += debts[pos];
res = Math.min(res, helper(debts, pos+1, count+1));
debts[i] -= debts[pos];//backtracking
}
}
return res;
}
}
Speed up:
有一点greedy + backtracking的感觉，首先用sort找出-5, 5这种pair, 然后从debts list中移除。
然后再用如上的backtracking方法
class Solution {
// DFS
public int minTransfers(int[][] transactions) {
Map<Integer, Long> map = new HashMap<>();
for(int[] t : transactions){
long val1 = map.getOrDefault(t[0], 0L);//balance
long val2 = map.getOrDefault(t[1], 0L);
map.put(t[0], val1-t[2]);
map.put(t[1], val2+t[2]);
}
List<Long> list = new ArrayList<>();
for(long val : map.values()){
if(val!=0) list.add(val);
}
int matchCount = removeMatch(list);
return matchCount + minTransStartFrom(list, 0);
}
private int removeMatch(List<Long> list) {
Collections.sort(list);
int left = 0;
int right = list.size() - 1;
int matchCount = 0;
while (left < right) {
if (list.get(left) + list.get(right) == 0) {
list.remove(left);
list.remove(right - 1);
right -= 2;
matchCount++;
} else if (list.get(left) + list.get(right) < 0) {
left++;
} else {
right--;
}
}
return matchCount;
}
private int minTransStartFrom(List<Long> list, int start) {
int result = Integer.MAX_VALUE;
int n = list.size();
while (start < n && list.get(start) == 0) {
start++;
}
if (start == n) {
return 0;
}
for (int i = start + 1; i < n; i++) {
if (list.get(i) * list.get(start) < 0) {
list.set(i, list.get(i) + list.get(start));
result = Math.min(result, 1 + minTransStartFrom(list, start + 1));
list.set(i, list.get(i) - list.get(start));
}
}
return result;
}
}
