Count and Say Problem in DSA Go - 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 Go

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 += string(count+'0') + string(prev[i-1])
            count = 1
        }
    }
    result += string(count+'0') + string(prev[len(prev)-1])
    return result
}

fmt.Println(countAndSay(4))

A 312211

B 111221

C 21

D 1211

💡 Hint

Trace the sequence from n=1 to n=4 step by step.

The Count and Say sequence starts with "1" at n=1.

n=2: "11" (one 1)

n=3: "21" (two 1s)

n=4: "1211" (one 2, two 1s)

❓ 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 Go

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 += string(count+'0') + string(prev[i-1])
            count = 1
        }
    }
    result += string(count+'0') + string(prev[len(prev)-1])
    return result
}

fmt.Println(countAndSay(5))

A 111221

B 312211

C 1211

D 211211

💡 Hint

Build from previous output n=4 = "1211".

n=4: "1211"

Count and say: one 1, one 2, two 1s → "111221"

❓ Predict Output

advanced

Output of Count and Say for n=6

What is the output of the Count and Say sequence for n=6?

DSA Go

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 += string(count+'0') + string(prev[i-1])
            count = 1
        }
    }
    result += string(count+'0') + string(prev[len(prev)-1])
    return result
}

fmt.Println(countAndSay(6))

A 312211

B 211211

C 13112221

D 111221

💡 Hint

Use output from n=5 = "111221" to generate n=6.

n=5: "111221"

Count and say: three 1s, two 2s, one 1 → "312211"

🧠 Conceptual

advanced

Understanding the Growth of Count and Say Sequence

Which statement best describes the length growth of the Count and Say sequence as n increases?

A The length remains constant after n=3.

B The length roughly doubles with each step.

C The length grows approximately by a factor of 1.3 each step.

D The length decreases as n increases.

💡 Hint

Consider how groups of digits are described in the sequence.

The length of the Count and Say sequence grows roughly by a factor of about 1.3 each step, not doubling or staying constant.

❓ Predict Output

expert

Output of Count and Say for n=7

What is the output of the Count and Say sequence for n=7?

DSA Go

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 += string(count+'0') + string(prev[i-1])
            count = 1
        }
    }
    result += string(count+'0') + string(prev[len(prev)-1])
    return result
}

fmt.Println(countAndSay(7))

A 211211

B 13112221

C 312211

D 1113213211

💡 Hint

Use output from n=6 = "312211" to generate n=7.

n=6: "312211"

Count and say: one 3, one 1, two 2s, two 1s → "13112221"

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