codility - TapeEquilibrium

문제 :

 

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].

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생각할 시간!!

 

1. 좌측값은 for문을 실행시켜 누적하면 해결. 그럼 우측값은??

2. 우측값은 총 합계 - 좌측값의 합계. 그렇기때문에 전체 합계를 구하기 위해 for문을 한번 실행. 총 2번의 for문이 실행되어야함. 

3. 원하는 결과는 좌측값 - 우측값의 절대값중 가장 작은값. 누적좌측값 - (총합계 - 누적좌측값) = 결과값. 

 

 

풀이 : 

class Solution {

    public int solution(int[] A) {
        
        int N = A.length;
        
        int min = 2000; // 초기 최소값은 배열값의 범위가 -1000 ~ 1000이기 때문에 |-1000-1000| 값인 2000으로 설정
        
        int totalSum = 0;
        int left = 0;
        
        // 총합계를 구한다.
        for(int cur : A) {
            totalSum += cur;
        }
        
        for(int P=1; P < N; P++) {
            
            left += A[P-1]; //누적 좌측값
            
            int cur = Math.abs(left - (totalSum - left)); // 결과값
            
            if(min > cur) { // 최소값과 비교
                min = cur;
            }    
        }
        
        
        return min;
    }
}