# Minimum Number of Transfers to Settle Debts - Google Top Interview Questions

### Problem Statement :

```You are given a two-dimensional list of integers transfers containing [source, dest, amount], which means that person source lent person dest a dollar amount equal to amount. Return the minimum number of person-to-person transfers that are required so that all debts are paid.

Constraints

n ≤ 13 where n is the number of participants

Example 1

Input

transfers = [

[0, 1, 50],

[1, 2, 50]

]

Output

1

Explanation

Person 0 gave person 1 \$50 and person 1 gave person 2 \$50.

So person 2 can directly give \$50 to person 0.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& transfers) {
map<int, int> M;
for (auto e : transfers) {
M[e[0]] -= e[2];
M[e[1]] += e[2];
}
vector<int> arr;
for (auto e : M) {
if (e.second) {
arr.push_back(e.second);
}
}
int n = arr.size();
int i, j;
vector<int> parts;
for (i = 1; i < (1 << n); ++i) {
int s = 0;
for (j = 0; j < n; ++j) {
if (i & (1 << j)) {
s += arr[j];
}
}
if (s == 0) {
parts.push_back(i);
}
}
vector<int> dp(1 << n, 1e9);
dp[0] = 0;
for (i = 1; i < (1 << n); ++i) {
for (int x : parts) {
if ((i & x) == x) {
dp[i] = min(dp[i], dp[i - x] + __builtin_popcount(x) - 1);
}
}
}
return dp[(1 << n) - 1];
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, transfers):
acc = Counter()

for u, v, val in transfers:
acc[u] += val
acc[v] -= val

@cache
def search(rem):

if len(rem) == 2:
if rem[0] == 0:
return 0
return 1

curr = rem[0]

if curr == 0:
return search(rem[1:])

best = 30

for i, val in enumerate(rem[1:]):
if (curr < 0 and val > 0) or (curr > 0 and val < 0):
next_state = tuple(
sorted(rem[1 : i + 1] + rem[i + 2 :] + (val + curr,), key=abs)
)
best = min(best, 1 + search(next_state))

return best

return search(tuple(sorted(acc.values(), key=abs)))```
```

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