### Tape Equilibrium

January 16, 2020

A non-empty 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:**

function solution(A);

that, given a non-empty 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.

Write an efficient algorithm for the following assumptions:

• 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].

Calculate all of the possible differences between the left and right parts.

```
function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
var smallest = Number.POSITIVE_INFINITY
var sumL=0, sumR=0, temp=0, i
for(i=0; i<A.length; i++) sumR += A[i] // total
for(i=1; i<A.length; i++){ //minimum 2 elements
sumL += A[i-1]
sumR -= A[i-1]
temp = Math.abs(sumL - sumR)
console.log("temp: ", temp)
if(temp < smallest) smallest = temp
}
//console.log("smallest: ", smallest)
return smallest
}
const A0 = [3, 1, 2, 4, 3]
console.log("smallest: ", solution(A0))
```