Count and Say Problem in DSA Go - Build from Scratch

Count and Say Problem

📖 Scenario: Imagine you are working with a simple communication system that encodes messages by counting consecutive digits and saying them out loud. This is called the "Count and Say" sequence.For example, starting with "1", the next term describes it as "one 1" which is "11". Then "11" is described as "two 1s" which is "21", and so on.

🎯 Goal: You will build a program that generates the nth term of the Count and Say sequence.This helps understand how to process strings and count repeated characters.

📋 What You'll Learn

Create a starting string for the sequence

Use a loop to generate terms up to n

Implement logic to count consecutive digits and build the next term

Print the final term after n iterations

💡 Why This Matters

🌍 Real World

The Count and Say sequence is a simple example of run-length encoding, a technique used in data compression to reduce repeated data.

💼 Career

Understanding string processing and loops is fundamental for software development, especially in tasks like parsing, encoding, and data transformation.

1

Create the starting string for the Count and Say sequence

Create a variable called current and set it to the string "1" as the first term of the sequence.

DSA Go

# Create a variable called current and set it to "1"
# Your code here

Hint

The first term of the Count and Say sequence is always "1".

2

Create a variable to hold the number of terms to generate

Create an integer variable called n and set it to 5 to generate the first 5 terms of the sequence.

DSA Go

package main

import "fmt"

func main() {
    current := "1"
    // Create an integer variable n and set it to 5
    // Your code here
}

Hint

Use n := 5 to set the number of terms.

3

Generate the Count and Say sequence up to the nth term

Use a for loop with a variable i from 1 to n-1 to generate the sequence. Inside the loop, create a variable next as an empty string. Use another loop to count consecutive characters in current and build next. After counting, set current = next.

DSA Go

package main

import "fmt"

func main() {
    current := "1"
    n := 5
    // Use a for loop from i := 1 to i < n
    // Inside, create next as empty string
    // Count consecutive characters in current
    // Build next string
    // Set current = next
    // Your code here
}

Hint

Use nested loops: outer for generating terms, inner for counting characters.

4

Print the nth term of the Count and Say sequence

Use fmt.Println(current) to print the final term stored in current after the loop.

DSA Go

package main

import "fmt"

func main() {
    current := "1"
    n := 5

    for i := 1; i < n; i++ {
        next := ""
        count := 1
        for j := 1; j < len(current); j++ {
            if current[j] == current[j-1] {
                count++
            } else {
                next += fmt.Sprintf("%d%c", count, current[j-1])
                count = 1
            }
        }
        next += fmt.Sprintf("%d%c", count, current[len(current)-1])
        current = next
    }
    // Print the final current string
    // Your code here
}

Hint

The 5th term of the Count and Say sequence is "111221".

Practice

1. What does the "Count and Say" sequence describe in simple terms?

easy

A. It describes the count of digits followed by the digit itself in the previous term.

B. It sums all digits of the previous term to get the next term.

C. It reverses the digits of the previous term to get the next term.

D. It replaces each digit with its square in the previous term.

Solution

1. Step 1: Understand the sequence rule

   The "Count and Say" sequence reads the previous term and counts consecutive digits, then says the count followed by the digit.

2. Step 2: Apply the rule to the sequence

   For example, starting with "1", the next term is "11" (one 1), then "21" (two 1s), and so on.

3. Final Answer:

   It describes the count of digits followed by the digit itself in the previous term. -> Option A

4. Quick Check:

   Count digits + say digit = It describes the count of digits followed by the digit itself in the previous term. [OK]

Hint: Remember: count digits, then say digit [OK]

Common Mistakes:

• Thinking it sums digits instead of counting
• Confusing it with reversing digits
• Replacing digits with their squares

2. Which of the following Go code snippets correctly initializes the first term of the Count and Say sequence as a string?

easy

A. var term string = "1"

B. term := 1

C. var term int = "1"

D. term := '1'

Solution

1. Step 1: Identify correct string initialization in Go

   In Go, strings are enclosed in double quotes "", so "1" is a string, while 1 is an integer.

2. Step 2: Check each option

   var term string = "1" uses var with string type and double quotes correctly. term := 1 assigns integer 1 without quotes. var term int = "1" tries to assign string to int type. term := '1' uses single quotes which are for rune, not string.

3. Final Answer:

   var term string = "1" -> Option A

4. Quick Check:

   String uses double quotes in Go = var term string = "1" [OK]

Hint: Use double quotes for strings in Go [OK]

Common Mistakes:

• Using single quotes for strings
• Assigning string to int variable
• Missing variable type or using wrong type

3. What is the output of the following Go code snippet for the Count and Say sequence when n=4?

package main
import "fmt"
func countAndSay(n int) string {
    if n == 1 {
        return "1"
    }
    prev := countAndSay(n - 1)
    result := ""
    count := 1
    for i := 1; i < len(prev); i++ {
        if prev[i] == prev[i-1] {
            count++
        } else {
            result += fmt.Sprintf("%d%c", count, prev[i-1])
            count = 1
        }
    }
    result += fmt.Sprintf("%d%c", count, prev[len(prev)-1])
    return result
}
func main() {
    fmt.Println(countAndSay(4))
}

medium

A. "21"

B. "111221"

C. "1211"

D. "312211"

Solution

1. Step 1: Trace the recursive calls for n=4

   n=1 returns "1"; n=2 reads "1" and returns "11" (one 1); n=3 reads "11" and returns "21" (two 1s); n=4 reads "21" and returns "1211" (one 2, one 1).

2. Step 2: Confirm output for n=4

   The output printed is "1211" which matches "1211".

3. Final Answer:

   "1211" -> Option C

4. Quick Check:

   Count and Say term 4 = "1211" [OK]

Hint: Trace terms step-by-step from 1 to n [OK]

Common Mistakes:

• Confusing term 3 output with term 4
• Missing last group count in loop
• Incorrect string concatenation

4. Identify the bug in the following Go code for Count and Say and select the fix:

func countAndSay(n int) string {
    if n == 1 {
        return "1"
    }
    prev := countAndSay(n - 1)
    result := ""
    count := 1
    for i := 0; i < len(prev); i++ {
        if i < len(prev)-1 && prev[i] == prev[i+1] {
            count++
        } else {
            result += fmt.Sprintf("%d%c", count, prev[i])
            count = 1
        }
    }
    return result
}

medium

A. No bug, code is correct

B. Remove the count variable and always append 1

C. Add a check to reset count only when digits differ

D. Change loop start from i=0 to i=1 and compare prev[i] with prev[i-1]

Solution

1. Step 1: Analyze the loop and comparisons

   The loop starts at i=0 and compares prev[i] with prev[i+1]. This causes out-of-range risk and incorrect counting because the first character has no previous to compare.

2. Step 2: Correct the loop to start at i=1 and compare with previous character

   Starting at i=1 and comparing prev[i] with prev[i-1] correctly counts consecutive digits without out-of-range errors.

3. Final Answer:

   Change loop start from i=0 to i=1 and compare prev[i] with prev[i-1] -> Option D

4. Quick Check:

   Loop index and comparison fix = Change loop start from i=0 to i=1 and compare prev[i] with prev[i-1] [OK]

Hint: Compare current with previous, start loop at 1 [OK]

Common Mistakes:

• Starting loop at 0 and comparing with next index
• Ignoring out-of-range errors
• Resetting count incorrectly

5. You want to generate the 5th term of the Count and Say sequence but only count digits '1' and '2' in the previous term, ignoring others. What is the 5th term output for this modified rule starting from "1"? Choose the correct output assuming the same counting and saying method but skipping digits other than '1' and '2'.

hard

A. "111221"

B. "1112"

C. "1211"

D. "312211"

Solution

1. Step 1: Generate normal Count and Say terms up to 5

   Normal terms: 1: "1", 2: "11", 3: "21", 4: "1211", 5: "111221".

2. Step 2: Apply modified rule ignoring digits other than '1' and '2'

   Term 4 is "1211". Counting only '1' and '2': "1" (one 1), "2" (one 2), "11" (two 1s). So count and say becomes "111211" (one 1, one 2, two 1s). But since we skip others, the correct modified 5th term is "111211".

3. Final Answer:

   "111211" -> Option B

4. Quick Check:

   Count only '1' and '2' digits = "111211" [OK]

Hint: Count only allowed digits, ignore others [OK]

Common Mistakes:

• Counting all digits as usual
• Including digits other than '1' and '2'
• Confusing normal and modified sequences