Codility Test (06. TapeEquilibrium)

Codility Test (06. TapeEquilibrium)

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:

class Solution { public int solution(int[] 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]. Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

問題回答

import java.util.stream.IntStream;

class Solution {
    public int solution(int[] A) {
        int firstTotal = 0;
        int lastTotal = 0;
        
        for(int i = 0; i < A.length; i++) {
            lastTotal += A[i];
        }
        
        int minResult = Integer.MAX_VALUE;
        
        for(int j = 0; j < A.length - 1; j++) {
            firstTotal += A[j];
            lastTotal -= A[j];
            int diff = Math.abs(firstTotal - lastTotal);
            if(minResult > diff) {
                minResult = diff;
            }
        }
        
        return minResult;
    }
}