CareerCup

public int solve(int[] array){
        if(array.length<=1){
            return 0;
        }

        int mod = (int)(1e9+7);
        int sum = 0;
        int pre_sum = array[0];
        for(int i=1;i<array.length;i++){
              sum += (pre_sum%array[i]);
              pre_sum+=array[i];
              sum %= mod;
              pre_sum %= mod;
        }

        int suffix_sum = array[array.length-1];
        for(int i=array.length-2;i>=0;i--){
            sum += (suffix_sum%array[i]);
            suffix_sum += array[i];
            sum %= mod;
            suffix_sum %= mod;
        }

        return sum;
    }

can you explain me the idea behind your solution

The solution is for sorted array.

Have anyone solved it ?

public class SumOfIntegers {
    public static void main(String[] args) {
        int[] arr = new int[] {1,2,3};
        long sum = 0;
        for(int i=0;i<arr.length;i++){
            int k= (i+1)%3;
            sum+=arr[i]%arr[k];
            sum+=arr[k]%arr[i];
        }

        sum+=sum%(Math.pow(10, 9)+7);

        System.out.println(sum);
    }

}

public class SumOfIntegers {
public static void main(String[] args) {
int[] arr = new int[] {1,2,3};
long sum = 0;
for(int i=0;i<arr.length;i++){
int k= (i+1)%3;
sum+=arr[i]%arr[k];
sum+=arr[k]%arr[i];
}

sum+=sum%(Math.pow(10, 9)+7);

System.out.println(sum);
}
}

import java.util.*;
public class Solution
{
public static void main(String[] args)
{
int[] A = new int[] {1, 2, 3};
int n = A[A.length-1];
long sum = 0;
for(int i = 1; i <= n; i++)
{
int q = n / i;
int r = n % i;
sum += (long)((i*(i-1)/2)*q + r*(r+1)/2);
}
System.out.println(sum % (long)(Math.pow(10, 9)+7));
}
}

import java.util.*;
public class Solution
{
public static void main(String[] args)
{
int[] A = new int[] {1, 2, 3};
int n = A[A.length-1];
long sum = 0;
for(int i = 1; i <= n; i++)
{
int q = n / i;
int r = n % i;
sum += (long)((i*(i-1)/2)*q + r*(r+1)/2);
}
System.out.println(sum % (long)(Math.pow(10, 9)+7));
}
}