What is the time complexity of assigning multiple elements to the array using the notation that uses only a single code line?

Question

I am referring to something like:

int[] array = {1, 2 , 3, 4, 5};

Where the assignment of a value to each array index is done "all in a single line of code".

I do not really know for sure what happens in the background with this, but what I have imagined is that it's something like an iteration through each index in the array, and then assigning the inputted int to that index:

for (int i = 0; i<array.length; i++){
array[i] = <given input>  //for array[1] the value will be 1, array[2] is 2 and so on...
}

If that is the case, it is in O(n) right? Though again, it is just my assumption of how the multiple assignments to an array occurs.

Answer 1

Very related, so check it out: Java: what's the big-O time of declaring an array of size n?

The resulting bytecode for initializing an array involves the newarray instruction, which is O(n) as each element is initialized to the default state before anything else. See: JVM Spec., page 576:

[...] Each of the elements of the new array is initialized to the defaultinitial value (§2.3, §2.4) for the element type of the array type.

What you are using here additionally is an array initializer, so let's check out the bytecode:

public class Main {
    public static void main(String[] args) {
        int[] array = {1, 2 , 3, 4, 5};
    }
}

... is compiled to ...

public class Main {
  public Main();
    Code:
       0: aload_0
       1: invokespecial #1                  // Method java/lang/Object."<init>":()V
       4: return

  public static void main(java.lang.String[]);
    Code:
       0: iconst_5
       1: newarray       int
       3: dup
       4: iconst_0
       5: iconst_1
       6: iastore
       7: dup
       8: iconst_1
       9: iconst_2
      10: iastore
      11: dup
      12: iconst_2
      13: iconst_3
      14: iastore
      15: dup
      16: iconst_3
      17: iconst_4
      18: iastore
      19: dup
      20: iconst_4
      21: iconst_5
      22: iastore
      23: astore_1
      24: return
}

Notice the newarray with iconst_5 passed to it. This belongs to the "new int[5] part" of the code, where iconst_5 is just 5 ("integer constant 5"). Ignore the dup and astore_1, they're not so important for the question.

Notice the iconst_0, iconst_1 and iastore block, which is repeated for other integer constants. iastore is the instruction to store a value in an array, where the first parameter is the index and the second parameter is the value. Two lines before the iastore are the arguments. So what you see is that the bytecode reads each of the elements in your array initializer and stores it in the array.

So what the one-liner does is actually very similar (or equivalent) to this:

public class Main {
    public static void main(String[] args) {
        int[] array = new int[5];
        array[0] = 1;
        array[1] = 2;
        array[2] = 3;
        array[3] = 4;
        array[4] = 5;
    }
}

As you can see, the one-liner you've shown is actually 2 * O(n) at bytecode-level.

Answer 2

The answer from @akuzminykh is nice and sufficient.

But what you are fishing for is compile time preparation of data.

static final String STR = "....very long string ...";

This string is compiled as UTF-8 string in the constant pool of the class. A java String will normally hold its data as array of UTF-16 (2 byte) chars, but the newest java versions can hold a byte array together with an encoding of it. So assume this to be the case for our string constant.

Then still the bytes need to be copied. Hence for n bytes still O(n). But it will be fast and (likely) on the native code side. A string of 1 character will be loaded faster than of 10 000 characters. But the real effect probably not measurable, not apt for optimisation.