Let's understand the question first
Given an array [2, 3, 4], and you need to product an array [3*4, 2*4, 2*3] => [12, 8, 6]
In other words, current element is removed and multiply the rest of elements.
It is easy to use two loops to solve the problem, but the runtime is $\mathcal{O}(n^2)$
How to solve the problem with runtime $\mathcal{O}(n)$, but you allow to use space $\mathcal{O}(n)$

Here is the trick with dynamic programming.

/**
* Multiply all integers except the current one
* No Division is allowed
* Runtime is O(n)
*
* [2, 3, 4] => [3*4, 2*4, 2*3] => [12, 8, 6]
* 
* [_, 3, 4] => 3*4
* [2, _, 4] => 2*4
* [2, 3, _] => 2*3
*
* 1. Use arr1[] and arr2[] as accumulators
* 2. arr1[0] and arr2[len-1] as initial values
* 3. arr1[i+1] = arr1[i]*arr[i]
* 4. [i]*[i] => [i+1]
*/ 
public static int[] multiply(int[] arr){
    if (arr == null){
        throw new IllegalArgumentException("arr must not be null.");
    }else{
        int len = arr.length;
        int[] acc1 = new int[len];
        int[] acc2 = new int[len];
        if(len > 1){
            // accumulators acc1[], acc2[] 
            acc1[0] = acc2[len - 1] = 1;
            for(int i=1; i < len; i++){
                acc1[i] = arr[i - 1]*acc1[i - 1];
                // inverse index
                acc2[len - 1 - i] = arr[len - i - (i-1)]*acc2[len - 1 - (i - 1)];
            }
            for(int i=0; i < len; i++)
                acc1[i] = acc1[i]*acc2[i];
        }
        return acc1;
    }
}