https://discuss.leetcode.com/topic/76158/11-liner-9ms-dfs-solution-detailed-explanation

Solution：

思路：
Time Complexity: O(N!) Space Complexity: O(N)

Solution Code：

class Solution {
    public int minTransfers(int[][] transactions) {
        Map<Integer, Long> map = new HashMap();
        // build id -> debt 
        for(int[] t: transactions){
            long val1 = map.getOrDefault(t[0], 0L);
            long val2 = map.getOrDefault(t[1], 0L);
            map.put(t[0], val1 - t[2]);
            map.put(t[1], val2 + t[2]);
        }        

        // ignore those who dont has debt 
        List<Long> list = new ArrayList();
        for(long val: map.values()) {
            if(val != 0) list.add(val);
        }
        
        // list -> array, just for faster calc & convenient
        Long[] debts = new Long[list.size()];
        debts = list.toArray(debts);
        
        // to counteract 抵消
        return helper(debts, 0 , 0);
    }

    // dfs(backtracking)
    int helper(Long[] debts, int pos, int count) {
        while(pos < debts.length && debts[pos] == 0) pos++;
        int res = Integer.MAX_VALUE;
        long pre = 0;
        
        for(int i = pos + 1; i < debts.length; i++) {
            // look for other debts with opposite sign to debt[pos]
            // pre: avoid same debt
            if(debts[i] != pre && debts[i] * debts[pos] < 0) {
                debts[i] += debts[pos];
                res = Math.min(res, helper(debts, pos + 1, count + 1)); // enter recursive: debts[pos] has been taken care of
                debts[i] = debts[i] - debts[pos]; // backtrack
                pre = debts[i];
            }
        }
      return res == Integer.MAX_VALUE ? count : res;
    }
}