Codility--- TapeEquilibrium

Task description

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).

Elements of input arrays can be modified.

Solution

 

Programming language used: Java

Total time used: 8 minutes

Code: 11:49:20 UTC, java, final, score:  100

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

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.lang.Math;
class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
         int N = A.length;
        int [] after = new int[N];
        after[N-1] = A[N-1];
        for(int i=N-2;i>=0;i--) {
            after[i] = after[i+1] + A[i];
        }
        int min = Math.abs(after[0] - after[1]*2);
        for(int P=2;P<N;P++) {
            int temp = Math.abs(after[0] - after[P]*2);
            if(min > temp) 
                min = temp;
        }
        
        return min;
    }
}