**TapeEquilibrium**

Minimize
the value |(A[0] + ... + A[P-1]) - (A[P] + ... + A[N-1])|.

A
non-empty zero-indexed array A consisting of N integers is given. Array A
represents numbers on a tape.

Any
integer P, such that 0 < P < N, splits this tape into two non-empty
parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The
difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P −
1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In
other words, it is the absolute difference between the sum of the first part
and the sum of the second part.

**For example**

consider
array A such that:

A[0] = 3

A[1] = 1

A[2] = 2

A[3] = 4

A[4] = 3

We
can split this tape in four places:

P
= 1, difference = |3 − 10| = 7

P
= 2, difference = |4 − 9| = 5

P
= 3, difference = |6 − 7| = 1

P
= 4, difference = |10 − 3| = 7

Write
a function:

class
Solution { public int solution(int[] A); }

that,
given a non-empty zero-indexed array A of N integers, returns the minimal
difference that can be achieved.

For
example, given:

A[0] = 3

A[1] = 1

A[2] = 2

A[3] = 4

A[4] = 3

the
function should return 1, as explained above.

**Assume that:**

N
is an integer within the range [2..100,000];

each
element of array A is an integer within the range [−1,000..1,000].

**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).

**Solution:**

class Solution {

public int solution(int[] A) {

int len = A.length;

if(len==1) {

return Math.

*abs*(A[0]);
} else if(len==2) {

return Math.

*abs*((A[0]-A[1]));
}

double sum = 0;

for(int val : A) {

sum = sum+val;

}

double leftSum = 0,
difference = 0, minimum = Integer.

*MAX_VALUE*;
for(int i=0; i<len-1; i++) {

leftSum = leftSum + A[i];

difference = sum - 2*leftSum;

minimum = Math.

*min*(Math.*abs*(difference), minimum);
}

return (int) minimum;

}

}

**Check my submission:**

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