WAP to Find equilibrium index in Unsorted Array of Size N, Efficiently

The 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

(The 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, here is a precise definition: The integer k is an equilibrium index of a sequence A[0],A[1]..,A[n-1] if and only if 0<= k and sum(A[0..(k-1)])=sum(A[(k+1)..(n-1)]). Assume the sum of zero elements is equal to zero.

Write a function
int equi(int A[], int n)

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

The problem can be solved by using various approaches, the most common being simply to follow the equilibrium definition:

int equi ( int A[], int n ) {
int k, m, lsum, rsum;
for(k = 0; k < n; ++k) {
lsum = 0; rsum = 0;
for(m = 0; m < k; ++m) lsum += A[m];
for(m = k + 1; m < n; ++m) rsum += A[m];
if (lsum == rsum) return k;
}
return -1;
}

Time Complexity o(N^2)
Space Complexity O(1)
Run Here https://ideone.com/fThsV

Unfortunately, this approach has two disadvantages:

* it fails on large input data sets, since the time complexity is O(n2)
* it fails on large input values (for example if the input array contains values like MIN/MAX_INT) due to the arithmetic overflows

The solution analysis will detect such problems in submitted code:

Optimized Solution

We can fix the first problem with a better algorithm, and the second problem with a better data-type (for example, using long long type instead of int for sum computations). The key observation for better running time is to update the left/right sums in constant time instead of recomputing them from the scratch.

int equi(int arr[], int n) {
if (n==0) return -1;
long long sum = 0;
int i;
for(i=0;i
long long sum_left = 0;
for(i=0;i long long sum_right = sum - sum_left - (long long) arr[i];
if (sum_left == sum_right) return i;
sum_left += (long long) arr[i];
}
return -1;
}


Using this solution, you can obtain a perfect answer:

To return all equilibrium index we can print instead of returning the single index

Time Complexity O(N)
Space Complexity O(1)
Run Here https://ideone.com/OKM8d