## Bloomberg LP Interview Question

Software Engineers**Country:**United States

**Interview Type:**In-Person

Let's assume that the array has at *least three entries* and that we pick numbers from the array *without replacement*.

The solution is trivial if the array only contains positive numbers. In fact even if there's one negative number in the array, the solution is:

`case_0 = prod( nlargest( 3, array ) )`

Here, nlargest can be implemented using a heap. In case there are two or more negative numbers in the array, we need to handle two distinct cases (due to the fact that two negative numbers can yield a large positive product).

```
case_1 = prod( nlargest( 3, array ) )
case_2 = prod( max( array ) * nsmallest( 2, array ) )
```

Thus, the final solution that handles all cases (including case_0 which is identical to case_1) in linear time is:

`max( case_1, case_2 )`

A specific implementation in Python might be:

```
from heapq import nsmallest, nlargest
from numpy import prod
from itertools import combinations
def max_prod3(arr):
case_1 = prod(nlargest(3, arr))
case_2 = max(arr) * prod(nsmallest(2, arr))
return max(case_1, case_2)
def test_max_prod3(arr):
# compare with brute-force solution
correct = max(prod(triplet) for triplet in combinations(arr, 3))
assert max_prod3(arr) == correct, "incorrect result for input {}".format(arr)
```

```
public int largestProduct3(int[] array) {
Arrays.sort(array);
int n = array.length;
int maxProduct;
//case> pick the last 3 numbers(when the array has only negative, or only positive numbers)
maxProduct = array[n - 1] * array[n - 2] * array[n - 3];
//case> pick 2 numbers from the left end and 1 number from the right end
maxProduct = Math.max(maxProduct, array[0] * array[1] * array[n - 1]);
return maxProduct;
}
```

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if the array is only positive integers you can do it in O(n)

int max1 = int.MinValue;

int max2 = int.MinValue;

int max3 = int.MinValue;

for(int i = 0; i <A.Length;i++)

{

if(A[i] > max1)

{

max3 = max2;

max2 = max1;

max1 = A[i];

}

if(A[i] > max2)

{

max3 = max2;

max2 = A[i];

}

if (A[i] > max3)

max3 = A[i];

}

Console.WriteLine(max1 * max2 * max3);

if the array is only positive integers, you can do it in O(n)

```
int max1 = int.MinValue;
int max2 = int.MinValue;
int max3 = int.MinValue;
for(int i = 0; i <A.Length;i++)
{
if(A[i] > max1)
{
max3 = max2;
max2 = max1;
max1 = A[i];
}
if(A[i] > max2)
{
max3 = max2;
max2 = A[i];
}
if (A[i] > max3)
max3 = A[i];
}
Console.WriteLine(max1 * max2 * max3);
```

If the array contains only positive integers then one can do it in O(n) ...

```
int max1 = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
int max3 = Integer.MIN_VALUE;
for(int i = 0; i <A.length;i++)
{
if(A[i] > max1)
{
max3 = max2;
max2 = max1;
max1 = A[i];
}
else if(A[i] > max2)
{
max3 = max2;
max2 = A[i];
}
else if (A[i] > max3)
max3 = A[i];
}
System.out.println("Maximum product : " + (max1 * max2 * max3));
```

```
void max3(int& m1, int& m2, int& m3){
int max = max(max(m1,m2),m3);
int min = min(min(m1,m2),m3);
m2 = (m1+m2+m3) - max - min;
m1 = max;
m3 = min;
}
int findMaxProduct(const vector<int>& v1){
int sz = v.size();
int m1 = v[0], m2 = v[1] , m3 = v[2];
max3(m1, m2, m3);
for(int i=3; i<sz; i++){
if(v[i] > m3){
m3 = v[i];
max3(&m1, &m2, &m3);
}
}
cout << m1 * m2 * m3;
}
```

Actually, since the question is about multiplying the three maximum numbers we can still achieve a O(n) for whatever numbers beings used by using additional data structure such as hash map which contains a maximum numbers as a key and frequency of that number as a value. Basically, the idea would be to insert maximums in a hash map with their frequency. if that frequency is greater than or equal 3, then we are done. If it is 2, we are missing the third maximum, if only 1, that means we are missing the other 2 maximums. So, the idea here is to iterate over the array 3 times which is O(3n).

if all is positive: a*b*c is largest if you pick the largest three numbers

if negative numbers are alowed, you have two cases, the product of the two smallest and the largest or the product the three largest

to find the three largest and two smallest numbers iterate over the array and pick this numers out (as if you would only scan for max or min). The thing is k-Selection and leads to O(n*k), but if k is constant, it's O(n) and if k is small it's reasonable efficient (among others due to caches on the CPU)

then calculate the two cases and return the max of them.

done.

if you need coaching... ;-)

Time complexity: O(N)

```
int MaxProd3(vector<int> const &a)
{
int max_prod3 = 0;
if (a.size() >= 3) {
int max_val = max(a[0], a[1]);
int max_prod2 = a[0] * a[1];
int min_val = min(a[0], a[1]);
int min_prod2 = a[0] * a[1];
max_prod3 = a[0] * a[1] * a[2];
for (int i = 2; i < a.size(); ++i) {
max_prod3 = max(max_prod3, a[i] * max_prod2);
max_prod3 = max(max_prod3, a[i] * min_prod2);
max_prod2 = max(max_prod2, max_val * a[i]);
max_prod2 = max(max_prod2, min_val * a[i]);
min_prod2 = min(min_prod2, max_val * a[i]);
min_prod2 = min(min_prod2, min_val * a[i]);
max_val = max(max_val, a[i]);
min_val = min(min_val, a[i]);
}
}
return max_prod3;
}
```

C++ solution

```
int product3highest(const int values[], size_t count) {
int highest[3] = { INT_MIN, INT_MIN, INT_MIN }; // keep sorted ascending
while (count--) {
if (values[count] > highest[0]) { // First check against lower value in highest[].
if (values[count] >= highest[2]) {
highest[0] = highest[1];
highest[1] = highest[2];
highest[2] = values[count];
} else if (values[count] >= highest[1]) {
highest[0] = highest[1];
highest[1] = values[count];
} else {
highest[0] = values[count];
}
}
}
return highest[0] * highest[1] * highest[2];
}
```

@aonecoding,

- NoOne May 03, 2017if you look into your own algorithm, I can see there is a solution in exact n. I dislike O, I like Theta. So yes, in Theta(n).