Count and Say Problem in DSA Go - Execution Trace

Concept Flow - Count and Say Problem

Start with '1'

↓

Read current string

↓

Count consecutive digits

↓

Build next string: count + digit

↓

Repeat for n times

↓

Output final string

The process starts with '1', then reads the string, counts consecutive digits, builds the next string by saying counts and digits, and repeats until the nth term.

Execution Sample

DSA Go

func countAndSay(n int) string {
    result := "1"
    for i := 1; i < n; i++ {
        result = nextSequence(result)
    }
    return result
}

func nextSequence(s string) string {
    // counts digits and builds next string
}

Generates the nth term of the count-and-say sequence by repeatedly building the next sequence from the current one.

Execution Table

Step	Operation	Current String	Count Digits	Built Next String	Visual State
1	Start	1	-	1	1
2	Count and say on '1'	1	one '1'	11	1 -> 1
3	Count and say on '11'	11	two '1's	21	2 -> 1
4	Count and say on '21'	21	one '2', one '1'	1211	1 -> 2 -> 1 -> 1
5	Count and say on '1211'	1211	one '1', one '2', two '1's	111221	1 -> 1 -> 1 -> 2 -> 2 -> 1
6	Count and say on '111221'	111221	three '1's, two '2's, one '1'	312211	3 -> 1 -> 2 -> 2 -> 1 -> 1
7	Stop	312211	-	-	Final output

💡 Reached the nth term (6), stop building next sequence.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
result	1	11	21	1211	111221	312211	312211
currentString	1	1	11	21	1211	111221	312211
countDigits	-	one '1'	two '1's	one '2', one '1'	one '1', one '2', two '1's	three '1's, two '2's, one '1'	-

Key Moments - 3 Insights

Why do we count consecutive digits instead of individual digits?

Because the sequence describes counts of groups of the same digit, not single digits separately. See step 3 in execution_table where '11' is counted as two '1's, not one '1' twice.

Why does the next string length vary and sometimes get longer?

Because the next string encodes counts and digits, so groups of digits become pairs like '21' for two '1's, increasing length. See step 5 where '1211' becomes '111221' which is longer.

What happens if n=1?

The function returns the initial string '1' immediately without building next sequences, as shown in step 1 of execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the 'result' after step 3?

A 11

B 21

C 1211

D 1

💡 Hint

Check the 'Built Next String' column at step 3 in execution_table.

At which step does the current string become '1211'?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Current String' column in execution_table.

If the initial string was '2' instead of '1', what would be the built next string at step 2?

A 12

B 11

C 21

D 22

💡 Hint

Counting one '2' would produce '12' as count and digit.

Concept Snapshot

Count and Say Problem:
- Start with '1' as the first term.
- For each next term, read previous term, count consecutive digits.
- Build next term by concatenating counts and digits.
- Repeat until nth term.
- Output the final string.

Full Transcript

The Count and Say problem generates a sequence of strings where each term describes the previous term by counting consecutive digits. Starting from '1', each next term is built by reading the current string, counting how many times each digit appears consecutively, and writing down the count followed by the digit. This process repeats until the nth term is reached. For example, starting with '1', the next term is '11' (one '1'), then '21' (two '1's), then '1211' (one '2', one '1'), and so on. The execution table shows each step with the current string, counts, and the next string built. Variables track the current string and counts at each step. Key moments clarify why counting consecutive digits is important and how the string length changes. The visual quiz tests understanding of the sequence progression and counting logic.

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