Count and Say Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the first term of the count and say sequence.

DSA C

char* countAndSay(int n) {
    if (n == 1) return [1];
    // rest of the code
}

Drag options to blanks, or click blank then click option'

A"11"

B"0"

C"1"

D"2"

Attempts:

3 left

💡 Hint

Common Mistakes

Returning "0" instead of "1" as the first term.

Using numeric 1 instead of string "1".

The first term in the count and say sequence is always "1".

2fill in blank

medium

Complete the code to iterate through the previous term string.

DSA C

for (int i = 0; i < strlen(prev); [1]) {
    // process characters
}

Drag options to blanks, or click blank then click option'

Ai--

Bi=0

Ci+=2

Di++

Attempts:

3 left

💡 Hint

Common Mistakes

Using i-- which causes infinite loop.

Skipping characters by incrementing by 2.

We increment i by 1 to move through each character in the string.

3fill in blank

hard

Fix the error in the condition to count consecutive characters.

DSA C

while (j < strlen(prev) && prev[j] == [1]) {
    count++;
    j++;
}

Drag options to blanks, or click blank then click option'

Aprev[j]

Bprev[i]

Cprev[i+1]

Dprev[j+1]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing prev[j] with itself causing infinite loop.

Using prev[i+1] which may go out of bounds.

We compare characters at j with the character at i to count repeats.

4fill in blank

hard

Fill both blanks to append count and character to the result string.

DSA C

sprintf(temp, "%d%c", [1], [2]);
strcat(result, temp);

Drag options to blanks, or click blank then click option'

Acount

Bprev[i]

Cj

Dcount + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Appending j instead of count.

Appending character at j instead of i.

We append the count and the character at position i to the result.

5fill in blank

hard

Fill all three blanks to update indices and prepare for next iteration.

DSA C

i = [1];
j = [2];
count = [3];

Drag options to blanks, or click blank then click option'

Aj

Bi + 1

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Resetting count to 0 causing incorrect counts.

Not updating i and j correctly causing infinite loops.

Set i and j to j to move forward, reset count to 1 for next group.

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