package com.code;public class Test03_3 {public static int solution(int[] A) {int size = A.length;if (size<2){return -1;}int [] rightSum = new int[size];rightSum[size-1] = A[size-1];for(int i=size-2;i>=0;i--){rightSum[i] = A[i]+rightSum[i+1];}int [] leftSum = new int[size];leftSum[0] = A[0];for(int i=1;i<size-1;i++){leftSum[i] = A[i]+leftSum[i-1];}int min = 2147483647;for(int i=0;i<size-1;i++){min = Math.min(min, Math.abs(leftSum[i]-rightSum[i+1]));}return min;}public static void main(String[] args) {int a [] = {3,1,2,4,3};System.out.println(solution(a));int b[] = {1,3};System.out.println(solution(b));}
}/**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] = 3A[1] = 1A[2] = 2A[3] = 4A[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] = 3A[1] = 1A[2] = 2A[3] = 4A[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.
*/