What is the expected output of the following code? data = 'abbabadaadbaccabc' print(data.count('ab', 1))

The correct answer is A. 2. str.count(sub, start) searches for non-overlapping occurrences of sub beginning at the given start index. With start=1, the search skips index 0, meaning the 'ab' at position 0 is excluded. The remaining string 'bbabadaadbaccabc' contains 'ab' at original indices 3 and 14…

Question

What is the expected output of the following code?
data = 'abbabadaadbaccabc'
print(data.count('ab', 1))

Options

How the community answered

(52 responses)

A
83% (43)
B
12% (6)
C
4% (2)
D
2% (1)

Explanation

str.count(sub, start) searches for non-overlapping occurrences of sub beginning at the given start index. With start=1, the search skips index 0, meaning the 'ab' at position 0 is excluded. The remaining string 'bbabadaadbaccabc' contains 'ab' at original indices 3 and 14 - exactly 2 occurrences, making A correct.

Why the distractors fail:

C (3) is the trap: calling data.count('ab') with no start argument returns 3 (finds all occurrences at indices 0, 3, and 14). Many test-takers miss that start=1 drops the first match.
B (4) and D (5) reflect overcounting errors, likely from misreading overlapping or adjacent characters like 'bb' or 'ba' as 'ab'.

Memory tip: Think of the start parameter as a bookmark - count('sub', n) tears off everything before page n and only counts within what's left. If the first match starts before your bookmark, it's gone.

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