## Goldman Sachs Interview Question

Applications Developers**Country:**India

```
def solve( my_list ){
// 0 is left partition, 1 is right partition
// a bit pattern of size(my_list) = n signifies the selection
// 0 1 0 1 0 --> pick the 0 for left, 1 for right
// 1 2 3 4 5 --> left : [1,3,5], right : [2,4]
// this is the bitmap trick
n = size(my_list)
N = 2 ** n
// infinity with null, lame!
min_diff = { 'value' : num('inf') , 'partitions' : null }
for ( my_num : [1:N] ){
string_rep = str(my_num,2) // in binary
// left pad 0
string_rep = '0' ** ( n - size(string_rep) ) + string_rep
// partition them using when 0 comes left, 1 comes right
#(left_items, right_items ) = partition ( [0:n] ) where { string_rep[$.o] == _'0' }
left_sum = sum ( left_items )
right_sum = sum ( right_items )
// this is pretty straight forward
diff = #|left_sum - right_sum |
if ( diff < min_diff.value ){
min_diff.value = diff
min_diff.partitions = [ left_items, right_items]
}
}
min_diff // return
}
l = [1,2,3,4,5]
md = solve( l )
println( md )
```

Here you go.

- NoOne August 21, 2017[ geeksforgeeks.org/partition-a-set-into-two-subsets-such-that-the-difference-of-subset-sums-is-minimum/ ]

I particularly did not like the dp thing. It won't work at all for any practical problems.