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@mindcruzer
Last active December 14, 2015 15:39
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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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