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.