codility-demo.java

/* 1. Equi
Find an index in an array such that its prefix sum equals its suffix sum.
Task description
This is a demo task. You can read about this task and its solutions in this blog post.
A zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e. 
A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.
For example, consider the following array A consisting of N = 8 elements:
  A[0] = -1
  A[1] =  3
  A[2] = -4   
  A[3] =  5
  A[4] =  1   
  A[5] = -6   
  A[6] =  2
  A[7] =  1
P = 1 is an equilibrium index of this array, because:
A[0] = −1 = A[2] + A[3] + A[4] + A[5] + A[6] + A[7]
P = 3 is an equilibrium index of this array, because:
A[0] + A[1] + A[2] = −2 = A[4] + A[5] + A[6] + A[7]
P = 7 is also an equilibrium index, because:
A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] = 0
and there are no elements with indices greater than 7.
P = 8 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.
Write a function:
class Solution { public int solution(int[] A); }
that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.
For example, given array A shown above, the function may return 1, 3 or 7, as explained above.
Assume that:
N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
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.
* */

// you can also use imports, for example:
// import java.util.*;

// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        int len = A.length;
        if (len < 1) {
            return -1; // cannot be anything, lol wat
        }
        if (len == 1) {
            return 0;
        }
        // all we look for is the indices where the sums are the same, these are the equilibrium points.
        long[] left = new long[len]; // test sums starting on  left;
        long[] right = new long[len]; // test sums starting on right;
        left[0] = A[0]; // first index?
        //System.out.println("A[len-1] = " + A[len-1]);
        right[len-1] = A[len-1]; // last index?
        for (int i=1; i<len; i+=1) {
            left[i] = left[i-1] + A[i];
        }
        for (int i=len-2; i>-1; i-=1) {
            right[i] = right[i+1] + A[i];
        }
        for (int i=0; i<len; i+=1) {
            //System.out.println("left at index [" + i + "]: " + (left[i]-A[i]) + "; right: " + (right[i]-A[i]));
            if ((left[i]-A[i]) == (right[i]-A[i])) return i;
        }
        return -1;
    }
}