Repeating a number based on position and value?

I am trying to learn about algorithmic thinking in my spare time from a book I got.

I am thinking about scenarios and trying to solve them using pseudocode in order to learn but I cannot approach this idea below. I am thinking of doing this with a nested loop but I don't know how to begin. Can you please assist?

If we have a 1d array that mentions how many times each position should be repeated for example: With N = 5, the 1d array is z = 2,1,1,3,2 Based on that, we should populate a new array with the times each number appears. Therefore, based on z we should have a 1d array A = 1,1,2,3,4,4,4,5,5.

Explanation: the first position of z has the number 2. Therefore, in the A array the number 1 should appear two times. continuing the number two of array z has the number 1, therefore the number 2 in the array A should appear only once, and so one for the rest numbers. So there is a pattern between the two arrays.

NOTE: This should not be using any functions or turning values to string and multiplying them. I am interested in the theoretical side of it and want to learn how to think for such problems.

UPDATE

Would it help if we have

an array with unsorted items:

 A: 2,3,1,2,4

and z where calculates how many times each position is mentioned.

z: 1,2,1,1

can we create a new array that will make the the content of A sorted based on the content of z?

Solution 1:

You can use list comprehension with nested loop:

result = [i+1 for i, freq in enumerate(z) for _ in range(freq)]

Without list comprehension:

result = []
for i, freq in enumerate(z):
    result.extend([i+1]*freq)

Without any functions (less Pythonic), using += and * operators for lists:

result = []
i = 1
for freq in z:
    result += [i] * freq
    i += 1

Without * operator for lists:

result = []
i = 1
for freq in z:
    while freq > 0:
        result += [i]
        freq -= 1
    i += 1

+= [i] is really a clumsy way to append(i), but at least it avoids the function, which you seem to be looking for.