[Codility] Lesson3. TapeEquilibrium - 문제풀이

Description

N개의 정수로 이루어진 배열 A가 비어 있지 않은 경우 얻을 수 있는 최소의 차이를 반환합니다.

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].

Solution 1. prefix sum

public static int solution(int[] A) {
    int len = A.length;
    int result = Integer.MAX_VALUE;
    int[] dp = new int[len];
    dp[0] = A[0];
    for(int i=1; i < len; i++){
        dp[i] = dp[i-1] + A[i];
    }

    for(int i=0; i < len-1; i++){
        int difference = Math.abs(dp[i]-(dp[len-1] - dp[i]));
        result = Math.min(difference,result);
    }

    return result;
}

prefixsum으로 각 요소까지의 합계를 저장하여 기억해 놓고 시작부터 현재 위치까지의 합계(dp[i]와 끝부터 현재다음까지의 합계의 차이를 계산해 나가고 가장 작은 값을 반환합니다.

Reference

TapeEquilibrium coding task - Learn to Code - Codility