tapeEquilibrium.js

/*

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].

*/

// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A) {
    // write your code in JavaScript (Node.js 8.9.4)
    
    if (A.length == 1) return Math.abs(A[0]);
    if (A.length == 2) return Math.abs(A[0] - A[1]);
    
    let p = [];
    
    //console.log('Before reduce: ' + A);
    
    let total = A.reduce((a, b) => a + b, 0);
    
    //console.log('After reduce: ' + A);
    
    let increasingTotal = 0;
    let reducedTotal = total;
    
    //let pElems = A.length - 1;
    let pElemsCounter = 0;
    
    for (let i = 0; i < A.length; i++) {
    
        let temp = A[i];
        
        //console.log('temp: ' + temp);
        
        increasingTotal = increasingTotal + temp;
        
        reducedTotal = reducedTotal - temp;
        
        //console.log('increasingTotal: ' + increasingTotal);
        //console.log('reducedTotal: ' + reducedTotal);
        //console.log('for storage :' + Math.abs(increasingTotal - reducedTotal))
        
        if (pElemsCounter < A.length - 1) {
            pElemsCounter++;
            p.push(Math.abs(increasingTotal - reducedTotal));
        }
    }
    
    //console.log(p);
    
    p = p.sort((a, b) => {
        return a - b;
    });
    
    //console.log('here: ' + p);
    
    return p[0];
}


//[-10, -20, -30, -40, 100]