Tape-Equilibrium Codility Training [closed]

Question

I received a codility test the other day for a job, as such I've been practicing using some of the problems from their training page Link

Unfortunately, I've only been able to get 83/100 on the Tape-Equilibrium question:

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.

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

Example: 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
In this case I would return 1 as it is the smallest difference.

N is an int, range [2..100,000]; each element of A is an int, range [−1,000..1,000]. It needs to be O(n) time complexity,

My code is as follows:

import java.math.*;
class Solution {
public int solution(int[] A) {
    
    long sumright = 0;
    long sumleft = 0;
    long ans;
    
    for (int i =1;i<A.length;i++)
        sumright += A[i];
    
    sumleft = A[0];
    ans =Math.abs(Math.abs(sumright)+Math.abs(sumleft));
    
    for (int P=1; P<A.length; P++)
    {
        if (Math.abs(Math.abs(sumleft) - Math.abs(sumright))<ans)
            ans = Math.abs(Math.abs(sumleft) - Math.abs(sumright));
        sumleft += A[P];
        sumright -=A[P];
    }
    return (int) ans;  
}

I went a bit mad with the Math.abs. The two test areas it fails on are "double" (which I think is two values, -1000 and 1000, and "small". http://codility.com/demo/results/demo9DAQ4T-2HS/

Any help would be appreciated, I want to make sure I'm not making any basic mistakes.

Answer 1

Your solution is already O(N). You need to remove the abs from sumleft and sumright.

if (Math.abs( sumleft - sumright ) < ans)
{
  ans = Math.abs( sumleft - sumright );
}

Also before the second for loop,

ans =Math.abs( sumleft - sumright );

It should work.

Answer 2

100%, in Javascript

var i, ll = A.length, tot = 0, upto = 0, min = Number.MAX_INT;

for (i=0; i<ll; i++) tot += A[i];

for (i=0; i<ll-1; i++)
{
    upto += A[i];
    var a1 = upto, a2 = tot - a1, dif = Math.abs(a1 - a2);
    if (dif < min)
         min = dif;
}

return min;

Answer 3

I found perfect solution for TapeEquilibrium by Cheng on Codesays. I translated it to Java for anybody who is curious about it. Cheng's solution hit 100% on Codility

    public int solution(int[] A) {

    // write your code in Java SE 7
    int N = A.length;

    int sum1 = A[0];
    int sum2 = 0;
    int P = 1;
    for (int i = P; i < N; i++) {
        sum2 += A[i];
    }
    int diff = Math.abs(sum1 - sum2);

    for (; P < N-1; P++) {
        sum1 += A[P];
        sum2 -= A[P];

        int newDiff = Math.abs(sum1 - sum2);
        if (newDiff < diff) {
            diff = newDiff;
        }
    }
    return diff;
}