Time complexity of tape equilibirum As per the instructions on this Tape Equlibrium-Codility

I've written the code

import java.util.Arrays;
public class TapeEqulibrium {
  int main(int[] array) {
    int minimumDiff = 100000;
    for (int i = 1; i &lt; array.length; i++) {
        int differnce = Math.abs(Arrays.stream(array, 0, i).sum() - Arrays.stream(array, i, array.length).sum());
        minimumDiff = differnce &lt; minimumDiff ? differnce : minimumDiff;
    }
    return minimumDiff;
   }
}


On submission, I have got correct answers for all test cases. But It says that the time complexity is O(n*n)and taking more than 6.00 seconds. while writing I thought that it falls under time complexity of O(n) or something (as I don't know much about that). So I need to know how to calculate the time complexity of this code.
You inner loop contains:

int differnce = Math.abs(Arrays.stream(array, 0, i).sum() - Arrays.stream(array, i, array.length).sum());


That does (i-0) operations for first sum and (array.length-i) for second, for a total of array.length-i+i-0=array.length. As the size of input is n=array.length that really gives O(n*n). Replacing for-loops with Streams doesn't improve complexity, only hides it a bit.