Searching and counting elements in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Counting occurrences in a list

What is the output of this code that counts how many times the number 3 appears in the list?

Python

numbers = [1, 3, 5, 3, 7, 3, 9]
count_3 = numbers.count(3)
print(count_3)

A 4

B 2

C 3

D 1

💡 Hint

Use the list method that counts how many times an item appears.

The list method count(x) returns how many times x appears in the list. Here, 3 appears three times.

❓ Predict Output

intermediate

Finding the index of an element

What will be printed by this code that finds the first position of the string 'apple' in the list?

Python

fruits = ['banana', 'apple', 'orange', 'apple']
index = fruits.index('apple')
print(index)

A 0

B 1

C 3

D 2

💡 Hint

The index method returns the first position where the item is found.

The first 'apple' is at position 1 (counting from zero).

❓ Predict Output

advanced

Counting elements with a condition

What is the output of this code that counts how many numbers in the list are greater than 5?

Python

nums = [2, 7, 4, 9, 5, 8]
count = sum(1 for n in nums if n > 5)
print(count)

A 3

B 4

C 2

D 5

💡 Hint

Sum 1 for each number that meets the condition n > 5.

Numbers greater than 5 are 7, 9, and 8, so count is 3.

❓ Predict Output

advanced

Finding the first matching element with next()

What will this code print? It finds the first even number in the list or returns -1 if none found.

Python

numbers = [1, 3, 5, 8, 10]
first_even = next((x for x in numbers if x % 2 == 0), -1)
print(first_even)

A 8

B 10

C -1

D 5

💡 Hint

next() returns the first item from the generator that meets the condition.

The first even number is 8, so it prints 8.

❓ Predict Output

expert

Counting unique elements with a condition

What is the output of this code that counts how many unique words in the list have length greater than 3?

Python

words = ['cat', 'dog', 'elephant', 'dog', 'bird', 'cat', 'elephant']
unique_words = set(words)
count = sum(1 for w in unique_words if len(w) > 3)
print(count)

A 1

B 3

C 4

D 2

💡 Hint

Convert list to set to get unique words, then count those longer than 3 letters.

Unique words are {'cat', 'dog', 'elephant', 'bird'}. Only 'elephant' and 'bird' have length > 3, so count is 2.

Practice

1. Which Python operator checks if an element exists in a list?

easy

A. in

B. count()

C. find()

D. exists()

Solution

1. Step 1: Understand the purpose of in

   The in operator checks if an element is present in a list or other collection.

2. Step 2: Compare with other options

   count() counts occurrences, find() and exists() are not valid list operators in Python.

3. Final Answer:

   in -> Option A

4. Quick Check:

   Use in to check membership [OK]

Hint: Use in to check presence quickly [OK]

Common Mistakes:

• Confusing count() with membership check
• Using non-existent methods like find()
• Trying to use exists() which is invalid

2. Which of the following is the correct syntax to count how many times the number 5 appears in list nums?

easy

A. count(nums, 5)

B. nums.count(5)

C. nums.count = 5

D. nums.count[5]

Solution

1. Step 1: Identify the correct method call

   To count occurrences, use the list method count() with the element as argument: nums.count(5).

2. Step 2: Check syntax of other options

   count(nums, 5) is invalid syntax, nums.count = 5 assigns a value incorrectly, and nums.count[5] is invalid indexing.

3. Final Answer:

   nums.count(5) -> Option B

4. Quick Check:

   Use list.count(element) to count [OK]

Hint: Use list.count(value) to count occurrences [OK]

Common Mistakes:

• Using function call syntax incorrectly
• Assigning instead of calling method
• Using square brackets instead of parentheses

3. What is the output of this code?

fruits = ['apple', 'banana', 'apple', 'cherry']
print(fruits.count('apple'))

medium

A. Error

B. 1

C. 3

D. 2

Solution

1. Step 1: Understand the list contents

   The list fruits contains 'apple' twice, 'banana' once, and 'cherry' once.

2. Step 2: Apply count() method

   fruits.count('apple') counts how many times 'apple' appears, which is 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Counting 'apple' in list = 2 [OK]

Hint: Count returns how many times item appears [OK]

Common Mistakes:

• Counting unique items instead of occurrences
• Expecting index instead of count
• Confusing count with length

4. Find the error in this code that tries to count how many times 10 appears in numbers:

numbers = [10, 20, 10, 30]
count = numbers.count[10]
print(count)

medium

A. Using square brackets instead of parentheses for count method

B. Variable name 'count' is reserved and cannot be used

C. List 'numbers' is not defined

D. Missing import for count function

Solution

1. Step 1: Identify method call syntax

   Methods in Python are called with parentheses, not square brackets. numbers.count[10] is invalid syntax.

2. Step 2: Correct the syntax

   It should be numbers.count(10) to count occurrences of 10.

3. Final Answer:

   Using square brackets instead of parentheses for count method -> Option A

4. Quick Check:

   Method calls need parentheses, not brackets [OK]

Hint: Use parentheses () to call methods, not brackets [] [OK]

Common Mistakes:

• Using [] instead of () for method calls
• Thinking count is a function needing import
• Assuming variable names are reserved

5. Given a list data = [0, 1, 2, 0, 3, 0, 4], which code snippet counts how many zeros are in the list and prints a message only if zeros exist?

hard

A. if 0 not in data:
print(f\"Zeros found: {data.count(0)}\")

B. print(f\"Zeros found: {data.count(0)}\")

C. if data.count(0) > 0:
print(f\"Zeros found: {data.count(0)}\")

D. if data.contains(0):
print(f\"Zeros found: {data.count(0)}\")

Solution

1. Step 1: Understand the goal

   We want to count zeros and print only if there is at least one zero.

2. Step 2: Analyze each option

   if 0 not in data:
       print(f\"Zeros found: {data.count(0)}\")

   checks if 0 is NOT in data, printing only when NO zeros -- incorrect.

   print(f\"Zeros found: {data.count(0)}\")

   always prints, even if count is zero.

   if data.count(0) > 0:
       print(f\"Zeros found: {data.count(0)}\")

   checks if count > 0 then prints the count. Correct and efficient.

   if data.contains(0):
       print(f\"Zeros found: {data.count(0)}\")

   uses invalid contains().

3. Step 3: Choose best option

   if data.count(0) > 0:
       print(f\"Zeros found: {data.count(0)}\")

   is correct, checking count once and printing only if zeros exist.

4. Final Answer:

   if data.count(0) > 0: print(f\"Zeros found: {data.count(0)}\") -> Option C

5. Quick Check:

   Check count > 0 before printing [OK]

Hint: Check count > 0 to confirm presence before printing [OK]

Common Mistakes:

• Using invalid method contains()
• Printing count without checking if zero exists
• Using not in which prints when zeros are absent