Count and Say Problem in DSA Go

Mental Model

We describe a number by counting how many times each digit appears in order, then say that count and digit out loud.

Analogy: Imagine reading a list of colored beads aloud by saying how many beads of each color you see in a row, like 'two red, one blue'.

1 -> 11 -> 21 -> 1211 -> 111221 -> null

Dry Run Walkthrough

Input: n = 5 (generate the 5th term in the count and say sequence)

Goal: Find the 5th term by repeatedly describing the previous term

Step 1: Start with '1' as the first term

1

Why: The sequence always starts with '1'

Step 2: Describe '1': one '1' -> '11'

11

Why: Count the digits and say them to get the next term

Step 3: Describe '11': two '1's -> '21'

21

Why: Two 1s become '21'

Step 4: Describe '21': one '2', one '1' -> '1211'

1211

Why: Say counts and digits in order

Step 5: Describe '1211': one '1', one '2', two '1's -> '111221'

111221

Why: Final term for n=5

Result:

111221

Annotated Code

DSA Go

package main

import (
	"fmt"
	"strconv"
)

func countAndSay(n int) string {
	if n == 1 {
		return "1"
	}
	prev := countAndSay(n - 1) // get previous term
	result := ""
	count := 1
	for i := 1; i < len(prev); i++ {
		if prev[i] == prev[i-1] {
			count++ // count repeated digits
		} else {
			result += strconv.Itoa(count) + string(prev[i-1]) // say count and digit
			count = 1 // reset count
		}
	}
	result += strconv.Itoa(count) + string(prev[len(prev)-1]) // last group
	return result
}

func main() {
	n := 5
	fmt.Println(countAndSay(n))
}

prev := countAndSay(n - 1) // get previous term

recursively get the previous term to build current term

if prev[i] == prev[i-1] { count++ } else { ... }

count consecutive same digits

result += strconv.Itoa(count) + string(prev[i-1])

append count and digit to result when digit changes

result += strconv.Itoa(count) + string(prev[len(prev)-1])

append last counted group after loop ends

Output

111221

Complexity Analysis

Time: O(m * n) because for each term up to n, we process the string of length m which grows roughly exponentially

Space: O(m) for storing the current term string, where m is length of the term

vs Alternative: Iterative approach can avoid recursion stack but time complexity remains similar due to string processing

Edge Cases

n = 1 (smallest input)

Returns '1' immediately without recursion

DSA Go

if n == 1 { return "1" }

n is large (e.g., 30)

String length grows large, may cause performance issues

DSA Go

No explicit guard; recursion depth and string size grow

When to Use This Pattern

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

Common Mistakes

Mistake: Not handling the last group of digits after the loop ends
Fix: Add the last counted group to the result after the loop

Mistake: Confusing count and digit order when appending to result
Fix: Always append count first, then the digit

Summary

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

Use when you need to build a sequence by describing previous terms in a count-and-say manner.

The key is to count consecutive digits and say them as count followed by digit to form the next term.

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