## Facebook Interview Question

Software Engineers**Country:**United States

**Interview Type:**In-Person

```
void printCoinsUpTo(const vector<int> &coins, int val) {
vector<int> result;
vector<int> idx(coins.size(), 0);
result.push_back(0);
for (int i = 1; i <= val; ++i) {
int minVal = INT_MAX;
for (int i = 0; i < coins.size(); ++i) {
minVal = min(minVal, coins[i] + result[idx[i]]);
}
result.push_back(minVal);
for (int i = 0; i < coins.size(); ++i) {
if (coins[i] + result[idx[i]] == minVal) {
idx[i]++;
}
}
}
for(int i = 1; i < result.size(); ++i) {
cout << result[i] << endl;
}
}
```

Here is my C++ backtracking solution.

In summary in given vector of coins, we pick one element and deduct the coin value from the given target until we reach 0 or negative number.

```
class Solution {
private:
int sum;
set<int> h;
public:
void combinationSum(vector<int>& coins, int target) {
sum=0;
dfs (candidates, 0, target);
for (auto t :h)
cout<<t<<" ";
return res;
}
void dfs (vector<int>& cand, int beg, int t) {
if(t<=0) {
return;
}
for(int i =beg ; i<cand.size();++i) {
if (cand[i]<=t) {
sum+= cand[i];
h.insert(sum);
dfs (cand, i, t-cand[i]);
sum-=cand[i];
}
}
}
};
```

time complexity : n * 'search and insert of set'

= n * (2 * log n)

-> n log n

or

time complexity : log(1) + log(2) + ... + log(n)

= log(n!)

n/2 log(n/2) <= log(n!) <= n log n

so

O(n log n)

```
void printCoinsSum(const vector<int>& coins, int max_sum) {
int start=coins[0];
for(int i=1;i<coins.size();i++) {
if(start>coins[i])
start=coins[i];
}
set<int> coins_sum;
coins_sum.insert(0);
for(int sum=start;sum<=max_sum;sum++) {
for(int i=0;i<coins.size();i++) {
/* new_sum = coins[i] + previous values
new_sum - coins[i] = previous values
check 'new_sum - coins[i]' is in previous values
*/
if(coins_sum.find(sum-coins[i])!=coins_sum.end()) {
coins_sum.insert(sum);
break;
}
}
}
set<int>::const_iterator it=coins_sum.begin();
++it;
for(;it!=coins_sum.end();it++)
cout<<*it<<endl;
}
```

With PHP.

There is 3 possibilities to combine three different coins. Number can be valid between 1 to 1000 when get 0 that divided by any of three combination.

```
function possibleCombinedSums($c1, $c2, $c3){
$combined1 = $c1 + $c2;
$combined2 = $c1 + $c3;
$combined3 = $c2 + $c3;
for($sum = 1;$sum <= 1000;$sum++){
if($sum%$combined1 == 0 || $sum%$combined2 == 0 || $sum%$combined3 == 0){
print $sum.'<br/>';
}
}
}
possibleCombinedSums(5,10,15);
```

```
printSums(int[] coins) {
Arrays.sort(coins)
int minCoin = coins[0];
boolean[] sums = new boolean[1000 - coins[0] + 1];
for (int i = 0; i < coins.length; i++) {
fillSums(sums, coins[i], minCoin);
}
for (int i = 0; i < sums.length; i++) {
if (sums[i]) {
system.out.println(i + minCoin);
}
}
}
fillSums(boolean[] arr, int coin, int minCoin) {
arr[coin - minCoin] = true;
for (int i = coin + 1; i - minCoin < arr.length; i++ ) {
if ( (i- coin - minCoin) >= 0 && arr[i - coin - minCoin] ) {
arr[i - minCoin] = true
}
}
}
```

```
sums = range(1,1001)
numbers = [10, 15, 55]
new_sums = list(filter((lambda x: x%10 == 0 or x%15 == 0 or x%55 == 0), sums))
print(new_sums)
```

```
public coins(int a, int b, int c) {
int step = gcd(a, b);
step = gcd(step, c);
int start = min(min(a, b), c);
for (int i = start; i < 1000; i += step) {
if (canFill(i, 0, a, b, c)) {
System.out.println(i);
}
}
}
private boolean canFill(int needed, int my, int a, int b, int c) {
if (my == needed) {
return true;
} else if (my > needed) {
return false;
} else {
return canFill(needed, my + a, a, b, c)
|| canFill(needed, my + b, a, b, c)
|| canFill(needed, my + c, a, b, c);
}
}
private int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
```

```
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;
public class Solution {
private PriorityQueue<Integer> priorityQueue = new PriorityQueue<>();
public ArrayList<Integer> intersect(List<Integer> denoms)
{
int sum = 0;
denoms.sort(Comparator.comparingInt(o -> o));
priorityQueue = new PriorityQueue<>();
ArrayList<Integer> result = new ArrayList<>();
while (sum < 1000) {
for (int i = 0; i < denoms.size(); ++i) {
priorityQueue.add(sum + denoms.get(i));
}
sum += priorityQueue.peek();
if (sum < 1000) {
result.add(sum);
}
}
return result;
}
}
```

Typical Dynamic Programming

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