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Question 45 of 97

Q.The function given below is written to accept a string s as a parameter and return the number of vowels appearing in the string. The code has certain errors. Observe the code carefully and rewrite it after removing all the logical and syntax errors. Underline all the corrections made. def CountVowels(s): c=0 for ch in range(s): if 'aeiouAEIOU' in ch: c=+1 return(ch)

Tamil Nadu DgeCBSE Class XII Board 2026Subjective· 2mImportance★★★★★
46% · 45/97 Questions
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The original code had four errors: incorrect string iteration, flawed vowel checking logic, an assignment typo instead of an increment, and returning the wrong variable. The corrected function iterates through the string, checks if each character is a vowel, increments a counter, and returns the final count.

The problem asks us to identify and correct logical and syntax errors in a Python function designed to count vowels in a string. Understanding how to iterate through strings, perform character comparisons, and correctly increment counters are key concepts here.

Let's break down the original code and identify each error.

def CountVowels(s):
    c=0
    for ch in range(s):
        if 'aeiouAEIOU' in ch:
            c=+1
    return(ch)

Concept and Intuition

The goal is to go through each character of the input string s, one by one. For each character, we need to determine if it's a vowel (either lowercase or uppercase). If it is, we increment a counter. After checking all characters, the final value of this counter is the result we want to return.

The errors in the provided code stem from misunderstandings of basic Python string iteration, membership testing, and arithmetic assignment.

Step-by-Step Correction

  1. Incorrect String Iteration:

    The line for ch in range(s): is incorrect. The range() function in Python expects an integer argument (or start, stop, step integers) to generate a sequence of numbers. When s (a string) is passed to range(), it will raise a TypeError. To iterate directly over the characters of a string, you simply use for ch in s:.

    • Correction: Change for ch in range(s): to for ch in s:.
  2. Flawed Vowel Checking Logic:

    The condition if 'aeiouAEIOU' in ch: attempts to check if the entire string 'aeiouAEIOU' is a substring within the single character ch. This will always evaluate to False because ch is a single character, and it cannot contain a multi-character string like 'aeiouAEIOU'. The intent is to check if the character ch itself is present within the set of vowels.

    • Correction: Change if 'aeiouAEIOU' in ch: to if ch in 'aeiouAEIOU':. This correctly checks if the character ch is one of the characters in the vowel string.
    Tip

    Using ch.lower() in 'aeiou' is another common and often cleaner way to check for vowels, as it converts the character to lowercase first, simplifying the vowel set. However, the problem's original approach of checking against both cases is also valid once corrected.

  3. Assignment Typo Instead of Increment: …

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