TapeEquilibrium.md

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:

function solution(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). Elements of input arrays can be modified.

function solution(A) {
    // write your code in JavaScript (Node.js 6.4.0)
    var sum1 = 0;
    var sum2 = A.reduce(function(acc, i) {
        return acc + i;
    });

    var minDiff = Number.POSITIVE_INFINITY;
    
    for (var p = 0; p < A.length - 1; p++)
    {
        sum1 += A[p];
        sum2 -= A[p];
        var diff = sum2 - sum1;
        minDiff = Math.min(minDiff, Math.abs(diff));
    }
    
    return minDiff;
}