Quote:
Equilibrium index of a sequence is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes. For example, in a sequence A:
A[0]=-7 A[1]=1 A[2]=5 A[3]=2 A[4]=-4 A[5]=3 A[6]=0

3 is an equilibrium index, because:
A[0]+A[1]+A[2]=A[4]+A[5]+A[6]

6 is also an equilibrium index, because:
A[0]+A[1]+A[2]+A[3]+A[4]+A[5]=0

(sum of zero elements is zero) 7 is not an equilibrium index, because it is not a valid index of sequence A.

If you still have doubts, this is a precise definition: the integer k is an equilibrium index of a sequence if and only if and .

Assume the sum of zero elements is equal zero. Write a function

int equi(int[] A);

that given a sequence, returns its equilibrium index (any) or -1 if no equilibrium indexes exist. Assume that the sequence may be very long.

def equi ( A ):    equi_index = -1    B = list()    # Construct the first two.     B.append({'prev_total':0, 'post_total':0, 'total':A[0]})    B.append({'prev_total':A[0], 'post_total':0, 'total':A[0]+A[1]})        # Loop through A, sans the first 2 values. This run will be for previous and current totals.      for i, v in enumerate(A[1:]):        prev_total = B[i-1]['prev_total'] + B[i-2]['prev_total']        total = v + B[i-1]['total']        record = {            'prev_total': prev_total,            'post_total':None,            'total': total            }         B.append(record)        # Now we can construct the last two for the post totals.    B[-1]['post_total'] = 0    B[-2]['post_total'] = A[-1]      # This run will set up the post totals      for i, v in enumerate(reversed(B[:-2])):        i = -3 - i # So we can efficiently access A in reversed order. A[-3] A[-4] etc.        B['post_total'] = B[i+1]['post_total'] + B[i+2]['post_total']        # Now we can perform our testing     for i, v in enumerate(B):        if v['prev_total'] == v['post_total'] and i != 0 and i != len(B):            return i        elif v['total'] == 0 and i != 0 and i!=len(B):            return i    return equi_index 

Quote: Original post by AndrewBC
Neat idea, but good lord their editor is crap.

Quote: Original post by way2lazy2care

Quote: Original post by AndrewBC
Neat idea, but good lord their editor is crap.

As I've gotten used to writing code in notepad, so the editor was fine for me.

I think if they gave the whole report with the code you had finished for the person to look at it would be fine.

Quote: Original post by Sneftel
Tried it out, also got the equilibrium index one (I wonder if that's their only sample). Not very difficult -- 100 out of 100 in about five minutes and ten lines of code -- but this is exactly the sort of first-pass filter that's useful for companies looking for programmers. Of course, any decent programmer should be able to come up with code tests, but this does save them the trouble, and allows them to grade code outside their language areas. Potentially very useful for employers.

Quote: Original post by AndrewBC
HM, now I feel like I'm missing out on something. How were you able to figure it up so quickly? Do you have any hints for ignoramus I?

def equi(A):    totals = list()    accum = 0    for element in A:        totals.append(accum)        accum += element    for idx in range(0, len(A)):        if totals[idx] == accum - totals[idx] - A[idx]:            return idx    return -1

Quote: Original post by crusadingknight
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Quote: Original post by Sneftel
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