public static void main(String[] args) { int[] returnVal = multiplyArrayItems(new int[]{1,2,3,4,5}); System.out.println(String.join(",", Arrays.stream(returnVal).mapToObj(String::valueOf).toArray(String[]::new))); } static int[] multiplyArrayItemsSE(int[] arr){ int[] returnArr = new int[arr.length]; ScriptEngineManager manager = new ScriptEngineManager(); ScriptEngine engine = manager.getEngineByName("js"); try { String[] sarr = Arrays.stream(arr).mapToObj(String::valueOf).toArray(String[]::new); int totalVal = (int) engine.eval[String.join("*", sarr)); for(int i = 0; i < arr.length; i++){ returnArr[i] = totalVal / arr[i]; } } catch (ScriptException e) { // TODO Auto-generated catch block e.printStackTrace(); } return returnArr; }

A O(2*N) = O(N) solution would include: 1. Get the product of all the numbers in the input array; 2. Set the ith number as the product, in step1, divided by the ith number in the input; Of course, the obvious solution would have complexity of O(n^2): get the new array ith number by getting the product of all the input numbers except for the ith position number.