Count and Say Problem in DSA C - 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 C

char* countAndSay(int n);

int main() {
    char* result = countAndSay(4);
    printf("%s\n", result);
    return 0;
}

// Assume countAndSay is implemented correctly

A "21"

B "111221"

C "312211"

D "1211"

💡 Hint

Remember the sequence starts with "1" and each term describes the previous term.

The sequence starts as: 1, 11, 21, 1211, ... For n=4, the output is "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 C

char* countAndSay(int n);

int main() {
    char* result = countAndSay(5);
    printf("%s\n", result);
    return 0;
}

// Assume countAndSay is implemented correctly

A "111221"

B "312211"

C "1211"

D "211211"

💡 Hint

Build from previous term n=4 which is "1211".

For n=5, describe "1211": one '1', one '2', two '1's → "111221".

🔧 Debug

advanced

Identify the error in Count and Say implementation

What error will this Count and Say code produce when n=3?

DSA C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* countAndSay(int n) {
    if (n == 1) return "1";
    char* prev = countAndSay(n - 1);
    int len = strlen(prev);
    char* result = malloc(len * 2 + 1);
    int count = 1, pos = 0;
    for (int i = 1; i <= len; i++) {
        if (prev[i] == prev[i - 1]) {
            count++;
        } else {
            result[pos++] = count + '0';
            result[pos++] = prev[i - 1];
            count = 1;
        }
    }
    result[pos] = '\0';
    return result;
}

int main() {
    char* res = countAndSay(3);
    printf("%s\n", res);
    free(res);
    return 0;
}

A Memory leak due to returning string literal in base case

B Incorrect output string for n=3

C Segmentation fault due to accessing prev[len] out of bounds

D Compilation error due to missing header

💡 Hint

Check the loop boundary and array indexing carefully.

The loop runs i from 1 to len inclusive, so prev[i] accesses prev[len] which is out of bounds causing segmentation fault.

🧠 Conceptual

advanced

Time complexity of Count and Say sequence generation

What is the time complexity of generating the nth term in the Count and Say sequence using the standard iterative approach?

A O(n * 2^n)

B O(n^2)

C O(2^n)

D O(n)

💡 Hint

Consider how the length of each term grows and how many terms are generated.

Each term roughly doubles in length, so length is about 2^(n). Generating each term requires processing the previous term fully, so total work is about n * 2^n.

🚀 Application

expert

Count and Say sequence: Find length of nth term

Given the Count and Say sequence, which approach efficiently computes the length of the nth term without generating the full string?

A Simulate the sequence generation and count characters each time

B Use a dynamic programming approach storing lengths only

C Use a mathematical closed-form formula for length

D Use a recursive formula based on previous term lengths

💡 Hint

Think about storing intermediate results to avoid full string construction.

Dynamic programming can store lengths of previous terms and compute next length by simulating counts without building strings, improving efficiency.

Practice

1. What is the output of the first term in the Count and Say sequence?

easy

A. "1"

B. "11"

C. "21"

D. "1211"

Solution

1. Step 1: Understand the first term definition

   The first term in the Count and Say sequence is always "1" by definition.

2. Step 2: Confirm the output

   Since the first term is "1", the output is exactly "1".

3. Final Answer:

   "1" -> Option A

4. Quick Check:

   First term = "1" [OK]

Hint: First term is always "1" in Count and Say [OK]

Common Mistakes:

• Confusing first term with second term
• Thinking sequence starts with "11"
• Mixing up output format

2. Which of the following is the correct syntax to generate the next term from a given term char* term in C for the Count and Say problem?

easy

A. Use a loop to count consecutive characters and build a new string

B. Directly print the term without processing

C. Use recursion without counting characters

D. Sort the characters in the term before counting

Solution

1. Step 1: Understand the Count and Say logic

   The next term is generated by counting consecutive characters and describing them.

2. Step 2: Identify correct approach

   Using a loop to count consecutive characters and build a new string matches the problem's logic.

3. Final Answer:

   Use a loop to count consecutive characters and build a new string -> Option A

4. Quick Check:

   Loop + count consecutive chars = correct syntax [OK]

Hint: Count consecutive chars with loop to build next term [OK]

Common Mistakes:

• Skipping counting step
• Trying to sort characters
• Using recursion without counting

3. What is the output of the Count and Say sequence at term 4?

medium

A. "21"

B. "111221"

C. "1211"

D. "3112"

Solution

1. Step 1: Generate terms up to 4

   Term 1: "1"
   Term 2: "11" (one 1)
   Term 3: "21" (two 1s)
   Term 4: "1211" (one 2, two 1s)

2. Step 2: Confirm term 4 output

   The fourth term is "1211" describing term 3.

3. Final Answer:

   "1211" -> Option C

4. Quick Check:

   Term 4 = "1211" [OK]

Hint: Build terms stepwise: count then say [OK]

Common Mistakes:

• Skipping terms and guessing output
• Mixing term 3 and 4
• Incorrect counting of digits

4. Given the following C code snippet for Count and Say, what is the error?

char* nextTerm(char* term) {
    int count = 1;
    char result[100];
    int index = 0;
    for (int i = 0; term[i] != '\0'; i++) {
        if (term[i] == term[i+1]) {
            count++;
        } else {
            result[index++] = count + '0';
            result[index++] = term[i];
            count = 1;
        }
    }
    result[index] = '\0';
    return result;
}

medium

A. Loop condition should be i < strlen(term)

B. Incorrect character comparison operator

C. Missing initialization of count variable

D. Returning pointer to local array causes undefined behavior

Solution

1. Step 1: Analyze return of local array

   The function returns pointer to local array 'result' which is destroyed after function ends.

2. Step 2: Identify undefined behavior

   Returning local array pointer leads to undefined behavior as memory is invalid outside function.

3. Final Answer:

   Returning pointer to local array causes undefined behavior -> Option D

4. Quick Check:

   Local array pointer return = undefined behavior [OK]

Hint: Never return pointer to local array in C [OK]

Common Mistakes:

• Ignoring local array lifetime
• Assuming loop condition is wrong
• Overlooking pointer return issues

5. How would you modify the Count and Say algorithm to handle very large terms efficiently in C?

hard

A. Store all terms in a global array for quick access

B. Use dynamic memory allocation and process terms iteratively

C. Sort each term before counting to reduce complexity

D. Use fixed-size arrays and recursion for each term

Solution

1. Step 1: Understand large term challenges

   Large terms require flexible memory to avoid overflow and efficient processing.

2. Step 2: Choose suitable approach

   Dynamic memory allocation allows resizing and iterative processing avoids stack overflow.

3. Final Answer:

   Use dynamic memory allocation and process terms iteratively -> Option B

4. Quick Check:

   Dynamic memory + iteration = efficient large term handling [OK]

Hint: Use malloc and loops for big terms, avoid recursion [OK]

Common Mistakes:

• Using fixed arrays causing overflow
• Recursion causing stack overflow
• Sorting terms breaks sequence logic