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""" | |
A zero-indexed array A consisting of N integers is given. An equilibrium | |
index of this array is any integer P such that 0 = P < N and the sum of | |
elements of lower indices is equal to the sum of elements of higher indices, i.e. | |
A[0] + A[1] + ... + A[P-1] = A[P+1] + ... + A[N-2] + A[N-1]. | |
Sum of zero elements is assumed to be equal to 0. This can happen if | |
P = 0 or if P = N-1. | |
For example, consider the following array A consisting of N = 7 elements: | |
A[0] = -7 A[1] = 1 A[2] = 5 | |
A[3] = 2 A[4] = -4 A[5] = 3 | |
A[6] = 0 | |
P = 3 is an equilibrium index of this array, because: | |
A[0] + A[1] + A[2] = A[4] + A[5] + A[6] | |
P = 6 is also an equilibrium index, because: | |
A[0] + A[1] + A[2] + A[3] + A[4] + A[5] = 0 | |
and there are no elements with indices greater than 6. | |
P = 7 is not an equilibrium index, because it does not fulfill the condition | |
0 = P < N. | |
Write a function | |
def equi(A) | |
that, given a zero-indexed array A consisting of N integers, returns any of its | |
equilibrium indices. The function should return -1 if no equilibrium index exists. | |
Assume that: | |
N is an integer within the range [0..10,000,000]; | |
each element of array A is an integer within the range | |
[-2,147,483,648..2,147,483,647]. | |
For example, given array A such that | |
A[0] = -7 A[1] = 1 A[2] = 5 | |
A[3] = 2 A[4] = -4 A[5] = 3 | |
A[6] = 0 | |
the function may return 3 or 6, as explained above. | |
Complexity: | |
expected worst-case time complexity is O(N); | |
expected worst-case space complexity is O(N), beyond input storage (not | |
counting the storage required for input arguments). | |
Elements of input arrays can be modified. | |
""" | |
def equi (A): | |
l = len(A) | |
s = sum(A) | |
ps = 0 | |
indicies = [] | |
for i in range(0, l): | |
ps += A[i - 1] if i > 0 else 0 | |
if (i == 0 and (s - A[i]) == 0) or ((s - ps - A[i]) == ps): | |
indicies.append(i) | |
return indicies if indicies else -1 | |
test = [-7, 1, 5, 2, -4, 3, 0] | |
print equi(test) |
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