## Wednesday, 2 December 2015

### Tape Equilibrium

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:
https://codility.com/demo/results/trainingJJUXQ9-2FK/