java - 优化数组的分区

Question

我正在解决一个编程挑战问题，但我的解决方案是给出大量的超时/错误。谁能帮我优化我的解决方案？

问题:

You are given an array A of N integers. Now you are required to fixed X such that the difference between the following two values is minimum:

  1. A[1] * A[2] * A[3] * ......... * A[X]
  2. A[X+1] * A[X+2] * ........... * A[N]

and if there is more value of X then print the smallest one.

Constraint:

• 1 <= 1 <= 10^5
• 1 <= A[i] <= 10^18

Input:

• The first line contains integer N (for size)
• The second line contains space separated numbers (for array)

import java.util.*;
public class Main
{
    public static void main(String[] args) {
        Scanner s=new Scanner(System.in)
        int size=Integer.parseInt(s.nextLine);
        long arr[]=new long[size];
        for(int i=0;i<=size;i++){
            arr[i]=s.nextLong();
        }   
        long part1=1,part2=1;
        long diff=1;long minIndex=0;long minNo=0;
        
        for(int k=0;k<size-1;k++){
            part1=1;part2=1;
            //minIndex=k;
            for (int i=0;i<=k ; i++){
                part1=part1*arr[i];
            } 
            for(int j=k+1;j<=size;j++){
                part2=part2*arr[j];
            }
            //System.out.println(part1+"---"+part2);
            diff=Math.abs(part1-part2);
            if(k==0){
                minNo=diff;
                minIndex=k;
            }
            //System.out.println(diff);
            if(minNo>diff){
                
                 minNo=diff;
                 minIndex=k;
            }
               
            
        }
        System.out.println("MinNo: "+minNo+" Index: "+minIndex);
        
        
        
    }
}

我正在测试这个输入

5
9090909090909009 780009090900909 898989898898898 98998 9999776765576765

答案应该是 2(如果从零开始计数，则为 1)但我的代码给出的是 4。

Answer 1

虽然@Mukesh Prajapati 建议的答案有效，但仍然有更快更好的方法来做到这一点。

您可以使用 log 来缩短值，这样您就可以在 log 计算中添加或减去值，因为现在加法意味着乘法，减法意味着分配。现在您的问题简化为在数组中找到一个点，其中左侧元素的总和最接近右侧元素。

您存储累积总和以便快速查找。这使您能够快速计算数组的左和和右和。最终的最小差异在 ans 中，而索引在 index 变量中。

void partition(int n, vector<double> &a) {
    double total = 0; vector<double> sum_array_a;

    for(auto &x: a) {
            x = log10(x);
            total += x;
            sum_array_a.push_back(total);
    }

    double ans = INFINITY, index = -1;

    for(int i = 0; i < n; i++) {    // Check for all points if you can split here
            double left = sum_array_a[i];
            double right = total - left;    // Right side sum of elements
            double diff = abs(left - right);
            if(diff < ans) {
                    ans = diff;
                    index = i;
            }
    }

    printf("%f", index);

}