Count and Say Problem in DSA C

Mental Model

We describe a number by counting groups of the same digit and saying the count followed by the digit.

Analogy: Imagine reading a line of colored beads and saying aloud how many beads of each color appear in a row before the color changes.

1
↓
"one 1" -> 11
↓
"two 1s" -> 21
↓
"one 2, one 1" -> 1211

Dry Run Walkthrough

Input: n = 4 (generate the 4th term starting from '1')

Goal: Find the 4th term in the count and say sequence starting with '1'

Step 1: Start with base term '1'

1

Why: The sequence always starts with '1'

Step 2: Read '1' as 'one 1', so next term is '11'

11

Why: We count one '1' and say '11'

Step 3: Read '11' as 'two 1s', so next term is '21'

21

Why: We count two '1's and say '21'

Step 4: Read '21' as 'one 2, one 1', so next term is '1211'

1211

Why: We count one '2' then one '1' and say '1211'

Result:

1211

Annotated Code

DSA C

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

// Function to generate the next term from current term
char* nextTerm(const char* term) {
    int len = strlen(term);
    char* result = (char*)malloc(len * 2 + 1); // max size
    int resIndex = 0;
    int count = 1;

    for (int i = 1; i <= len; i++) {
        if (i < len && term[i] == term[i - 1]) {
            count++;
        } else {
            // Append count and digit to result
            resIndex += sprintf(result + resIndex, "%d%c", count, term[i - 1]);
            count = 1;
        }
    }
    result[resIndex] = '\0';
    return result;
}

// Function to generate nth term of count and say sequence
char* countAndSay(int n) {
    if (n == 1) {
        char* base = (char*)malloc(2);
        strcpy(base, "1");
        return base;
    }
    char* prev = countAndSay(n - 1);
    char* curr = nextTerm(prev);
    free(prev);
    return curr;
}

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

if (i < len && term[i] == term[i - 1]) {

count consecutive same digits to group them

resIndex += sprintf(result + resIndex, "%d%c", count, term[i - 1]);

append count and digit to build next term

char* prev = countAndSay(n - 1);

recursively get previous term

char* curr = nextTerm(prev);

generate current term from previous

free(prev);

free memory of previous term to avoid leaks

Output

1211

Complexity Analysis

Time: O(m * n) because each term generation scans the previous term of length m, repeated n times

Space: O(m * n) due to storing strings for each term up to n

vs Alternative: Iterative approach can reduce recursion overhead but time complexity remains similar

Edge Cases

n = 1 (base case)

Returns '1' immediately without recursion

DSA C

if (n == 1) {

term with all identical digits (e.g., '111')

Counts all digits as one group correctly

DSA C

if (i < len && term[i] == term[i - 1]) {

term with alternating digits (e.g., '1212')

Counts each digit group separately

DSA C

else { resIndex += sprintf(result + resIndex, "%d%c", count, term[i - 1]); count = 1; }

When to Use This Pattern

When asked to generate a sequence where each term describes the previous term's digit groups, use the Count and Say pattern to build terms step-by-step.

Common Mistakes

Mistake: Not resetting count after appending a group
Fix: Reset count to 1 after appending count and digit

Mistake: Off-by-one error in loop causing missing last group
Fix: Loop until i <= length and handle last group in else block

Summary

Generates a sequence where each term describes the count of digits in the previous term.

Use when a problem requires describing groups of repeated digits in a sequence.

The key is to group consecutive digits and say their count followed by the digit.

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