Count and Say Problem in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Count and Say for n=4

What is the output of the Count and Say sequence for n=4?

DSA Python

def count_and_say(n):
    if n == 1:
        return "1"
    prev = count_and_say(n - 1)
    result = ""
    count = 1
    for i in range(1, len(prev)):
        if prev[i] == prev[i - 1]:
            count += 1
        else:
            result += str(count) + prev[i - 1]
            count = 1
    result += str(count) + prev[-1]
    return result

print(count_and_say(4))

A 1211

B 111221

C 21

D 312211

💡 Hint

Trace the sequence from n=1 to n=4 step by step.

The Count and Say sequence starts with "1" for n=1.

For n=2, it reads "1" as "one 1" → "11".

For n=3, it reads "11" as "two 1s" → "21".

For n=4, it reads "21" as "one 2, then one 1" → "1211".

❓ Predict Output

intermediate

Output of Count and Say for n=5

What is the output of the Count and Say sequence for n=5?

DSA Python

def count_and_say(n):
    if n == 1:
        return "1"
    prev = count_and_say(n - 1)
    result = ""
    count = 1
    for i in range(1, len(prev)):
        if prev[i] == prev[i - 1]:
            count += 1
        else:
            result += str(count) + prev[i - 1]
            count = 1
    result += str(count) + prev[-1]
    return result

print(count_and_say(5))

A 1211

B 211211

C 312211

D 111221

💡 Hint

Build the sequence from n=1 up to n=5 carefully.

From previous steps:

n=4 is "1211".

Reading "1211": one 1, one 2, two 1s → "111221".

🧠 Conceptual

advanced

Understanding the Growth of Count and Say Sequence Length

Which statement best describes how the length of the Count and Say sequence changes as n increases?

A The length roughly doubles with each step.

B The length grows approximately by a factor of 1.3 each step.

C The length stays the same for all n.

D The length decreases as n increases.

💡 Hint

Think about how the sequence describes groups of digits.

The length of the Count and Say sequence grows but not doubling each time.

It grows roughly by a factor around 1.3 per step due to grouping digits.

🔧 Debug

advanced

Identify the Error in Count and Say Implementation

What error will this code produce when calling count_and_say(3)?

DSA Python

def count_and_say(n):
    if n == 1:
        return "1"
    prev = count_and_say(n - 1)
    result = ""
    count = 1
    for i in range(len(prev)):
        if prev[i] == prev[i - 1]:
            count += 1
        else:
            result += str(count) + prev[i - 1]
            count = 1
    result += str(count) + prev[-1]
    return result

print(count_and_say(3))

A No runtime error, produces wrong output

B TypeError

C ValueError

D IndexError

💡 Hint

Look at the loop index starting point and how prev[i - 1] is accessed.

The loop starts at i=0, so when i=0, prev[i-1] becomes prev[-1], which is valid in Python (last character).

However, this leads to incorrect logic because it compares the first character to the last, instead of grouping consecutive digits properly.

The code runs without any runtime error but produces incorrect output ("11121" instead of "21" for n=3).

Fix: Change the loop to for i in range(1, len(prev)).

🚀 Application

expert

Count and Say Sequence - Number of Digits in 10th Term

How many digits are in the 10th term of the Count and Say sequence?

A 100

B 50

C 20

D 30

💡 Hint

Use the approximate growth factor of 1.3 per step starting from length 1 at n=1.

Length at n=1 is 1.

Length grows roughly by 1.3 times each step.

Length at n=10 ≈ 1 * (1.3^9) ≈ 10.6 (approximation is low).

Actual length is 20 digits at n=10.