Why Does Bubble Sort Use range(1, len(L)) Instead of Starting at 0?

[Rajeev Bagra]

When beginners learn Python loops, one common question appears:

for j in range(1, len(L)):

Why does this start at 1?

Isn’t Python like C, where indexing starts at 0?

Short Answer

Yes — Python lists are zero-indexed, just like C arrays.

But Bubble Sort starts this loop at 1 because the algorithm compares:

L[j - 1] and L[j]

Starting at 1 ensures the first comparison is:

L[0] and L[1]

which is correct.

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Python Lists Still Start at 0

Example:

L = [8, 4, 1, 6]

Indexes:

0 → 8
1 → 4
2 → 1
3 → 6

So Python definitely uses zero-based indexing.

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Why Not Start at 0?

The Bubble Sort code uses:

if L[j - 1] > L[j]:

If the loop started with:

j = 0

Then Python would calculate:

L[j - 1] = L[-1]

And in Python:

L[-1]

means the last element of the list.

So Python would compare:

last item vs first item

That is incorrect for Bubble Sort.

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Why Starting at 1 Works

First loop value:

j = 1

Then:

L[j - 1] = L[0]
L[j] = L[1]

So first comparison becomes:

first item vs second item

Perfect.

Next:

j = 2 → compare L[1] and L[2]
j = 3 → compare L[2] and L[3]

Exactly what Bubble Sort needs.

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Even in C, This Is Common

In C, if comparing:

arr[j - 1] > arr[j]

Many programmers also start:

for (j = 1; j < n; j++)

So this is not a Python-only thing.

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Alternative Python Version Starting at 0

You can rewrite Bubble Sort like this:

for j in range(0, len(L) - 1):
    if L[j] > L[j + 1]:
        swap...

Now comparisons become:

L[0] vs L[1]
L[1] vs L[2]
L[2] vs L[3]

Also correct.

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Mental Model

If code compares:

L[j - 1] and L[j]

Start loop at:

1

If code compares:

L[j] and L[j + 1]

Start loop at:

0

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One-Line Takeaway

Python lists start at 0, but Bubble Sort uses range(1, len(L)) so L[j-1] safely points to the previous item.

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